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BY THE SAME A UTHOR. 



ALGEBRA FOR COLLEGES. 

KEY TO ABOVE. 

ELEMENTARY ALGEBRA FOR 
SCHOOLS. Ready in July, 1882. 

ELEMENTS OF GEOMETRY. 

ELEMENTS OF PLANE AND SPHER- 
ICAL TRIGONOMETRY. 

LOGARITHMIC AND OTHER MATH- 
EMATICAL TABLES. 

TRIGONOMETRY AND TABLES. In 

one volume. 

ASTRONOMY. For Students and General 
Readers. By Simon Newcomb and Ed- 
ward S. Holden. 



JYEWCOMB'S MATHEMATICAL COURSE 



ELEMENTS 



OF 



PLANE AND SPHERICAL 
TRIGONOMETRY 



BY 



SIMON NEWCOMB 

Professor of Mathematics, United States Navy 




NEW YORK 
HENRY HOLT AND COMPANY 

1882 






Copyright, 1882, 

BY 

Henry Holt & Co. 




Electrotyped and Printed by 

S. W. GREEN'S SON, 

74 and 76 Beekman Street, 

New York. 






PEEFAOE. 



The distinctive features of the following work belong partly 
to the course of which it forms a part, and need but a brief 
statement. 

I. The device by which mathematical teaching is to be most 
promoted is, the author conceives, to be found in the minute sub- 
division of subjects, and the drill of the student in the separate 
details before combining them into a whole. The system to which 
we are thus led is seen in the arrangement of Chapters I., II., and V. 

II. By exercises in which the subject is taken up in a concrete 
form, the formation of mathematical conceptions is greatly facili- 
tated. An application of this principle is seen in the cases where 
the student is exercised in finding the values of trigonometric func- 
tions by construction and measurement. 

III. The problems for exercise are quite varied in their charac- 
ter, and are intended to test not only the student's knowledge of 
the usual methods of computation, but his ability to grasp them 
and trace them out in the numerous forms they may assume in 
practical applications. 

IV. In the arrangement, strictly logical order has been sub- 
ordinated to order of teaching. In accordance with this principle, 
all the simpler applications of the trigonometric functions have 
been disposed of before their complex relations. 

V. The scope of the work is generally limited to the subjects 
and treatment necessary in the fullest course of mathematics 
usually taught in our colleges and technological schools. The 
concluding chapter of each part perhaps exceeds the limit thus 



iv PREFACE. 

set. That of Part I. is an introduction to the employment of 
imaginary quantities in trigonometric developments, while that of 
Part II. is an introduction to the higher forms of solid geometry. 
YI. To the usual list of subjects treated, has been added t 
chapter on the theory of polygons. This theory is closely con 
nected with a variety of subjects, including geometry, quaternions, 
mechanics, graphical statics, surveying, and navigation, and there- 
fore deserves a more prominent place than has hitherto been 
assigned to it. 



TABLE OF CONTENTS, 



PART I. PLANE TRIGONOMETRY. 

PAGE 

Chapter I. Of Goniometry, or the Measure op Angles 1 

Angular measure, 1. The division of angles, 6. Natural measure 
of angles, 8. 

Chapter II. Op the Trigonometric Functions 11 

The sine, tangent, and secant, 11. Functions of unlimited angles, 15. 
The cosine, cotangent, and cosecant, 19. Values for the second, third, 
and fourth quadrants, 22. Special values of the trigonometric func- 
tions, 26. Angles corresponding to given functions, 27. Relations 
between the six trigonometric functions, 30. 

Chapter III. Of Right Triangles 36 

Fundamental relations, 36. Examples and Exercises in Expres- 
sion, 37. Solution of right triangles, 38. 
Chapter IV. Relations between Functions of Several Angles. . 44 
The addition and subtraction theorems, 44. Miscellaneous rela- 
tions, 47. Dimidiation, 49. 

Chapter V. Trigonometric Problems 53 

Chapter VI. Solution of Triangles in General 61 

Relations between the angles, 61. Case I. Given the angles and 
one side, 62. Case II. Given two sides and the angle opposite one of 
them, 64. Case III. Given the three sides, 66. Case IV. Given two 
sides and the included angle, 70. Areas of triangles, 73. 

Chapter VII. The Theory of Polygons 75 

Co-ordinates of a point, 75. Definition of direction, 77. Projections 
of lines, 78. Algebraic addition of lines, 80. Theorems of projections 
of the sides of a polygon, 81. Areas of polygons, 86. 

Chapter VIII. Trigonometric Developments . 91 

Development of sines and cosines in powers of the angle, 92. Imagi- 
nary exponentials, 96. Sines and cosines of multiple arcs, 97. Powers 
of the cosine, 99. Powers of the sine, 101. Trigonometric forms of 
imaginary expressions, 104. Roots of unity, 108. 



vi CONTENTS. 

PART II. SPHERICAL TRIGONOMETRY. 

PAGE 

Chapter I. Fundamental Principles 113 

Preliminary notions, 113. Fundamental equations, 116. Polar tri- 
angles, 120. Forms of the fundamental equations, 123. 
Chapter II. Right and Quadrantal Triangles 129 

Fundamental definitions and theorems, 129. Formulae for right 
triangles, 130. Napier's rules, 132. Relations among four parts, 133. 
Quadrantal triangles, 137. 
Chapter III. Transformation op the Formulae op Spherical Trigo- 
nometry 139 

Expressions when the three sides or the three angles are given, 140. 
Cagnoli's equation, 143. Gauss's equations, 144. Napier's analogies, 
146. 
Chapter IV. Miscellaneous Applications 148 

Geographical applications, 148. Geometrical applications, 150. Rec- 
tangular lines and planes, 150. Direction of a line in space, 154. 
Relation to latitude and longitude, 156. Position of a point, 156. 
Polar distance and longitude, 157. Rectangular co-ordinates, 157. 
Projection of one line upon another, 161. Plane triangles in space, 
162. 



Elements of Trigonometry 



PART I. 
PLANE TRIGONOMETRY. 



CHAPTER I. 

OF G0N10METRY, OR TEE MEASURE OF ANGLES. 



1. Definition. Trigonometry is that branch of geometry 
in which the relations of lines and angles are treated by algebraic 
methods. 

2. Def. An angle is the figure formed by two straight lines 
emanating from the same point, called the vertex of the angle. 

Def. The lines which form an angle are called its sides, 

3. Measures of Angles. An angle is measured by the length 
of a circular arc having its centre at the 
vertex of the angle and its ends on the sides 
of the angle. / 

If the angle to be measured is AOB, / 
we conceive that with an arbitrary radius j 
Oa an arc is drawn from a to h. \ 

"We regard as the positive direction that j|\ / 

in which the arc is described by a motion 

opposite to that of the hands of a watch, and as the negative direc- 
tion that in which the hands move. 

Hence we may consider the angle as measured either by the 




2 PLANE TRIGONOMETRY. 

arc ah considered as positive, or the conjugate arc aMb considered 
as negative. The numerical sum of these two arcs is equal to a 
circumference. 

As an example of the use of algebraic signs, we may mention 
their application to the latitude of places to distinguish them as 
north and south. Thus, a city in 42° north latitude is said to have 
a latitude of -f- 42°, and one 42° south of the equator is said to be 
in latitude — 42°. 

The absolute length of the arc will depend not only upon 
the magnitude of the angle, but upon the radius with which the 
arc is drawn. To avoid ambiguity from this cause, the unit of 
arc is supposed to be some fixed fraction of the circumference, 
and therefore greater the greater the radius. The arc is then 
indicated by the number of units and parts of a unit which it con- 
tains, and this number is the same for the same angle whatever 
the radius may be. 

To indicate the angle corresponding to any arc we call it the 
angle of the arc, or, for brevity, the angle-arc. 

4, The Sexagesimal Division, The following is the usual 
division : 

The circumference is divided into 360 units, called degrees ; 
Each degree is divided into 60 minutes ; 
Each minute is divided into 60 seconds. 
Then 

1 circumference = 360° = 21 600' = 1 296 000"; 

1 quadrant or right angle = 90°. 
This is called the sexagesimal division of the circle. 

5. The Centesimal Division. The sexagesimal division of the 
circle is by no means so convenient as one in which each unit is 
10 times or 100 times greater than the next smaller unit. The 
centesimal division was introduced by the French geometers at 
the time of the Re volution. In this system 

The circumference is divided into 400 grades ; 
The grade is divided into 100 minutes / 
The minute is divided into 100 seconds. 



OF GONIOMETRT. 

Hence 1 circumference = 400 grades = 40 000 min. = 4 000 000 
sec. ; which is commonly written 

400* r - = 40 000' = 4 000 000". 
Notwithstanding its greater convenience, this system never 
came into general use, owing to the difficulty of changing all the 
mathematical tables to correspond with it. 

6. Decimals of Degrees or Minutes. Sometimes, instead of 
seconds, decimals of a minute are used. Both minutes and seconds 
may be dispensed with and decimals of a degree be used in their 
place. 

"7, General Measure of an Angle. The best way of thinking 
of angular measure is to conceive the side OA of the angle to turn 
round on O until it reaches the position OB. In thus turning, a 
point A upon it will describe the circular arc which measures the 
angle A OB. The length of this arc will then be proportional to 
the amount by which OB turns in passing from OA to OB. 

The side may pass from OA to OB not only by describing 
the arc ab, but by moving through a whole revolution plus the arc 
a b, or through any number of revolutions plus the arc ab. When 
we consider the angle in the most general way, all these motions 
will equally measure the angle. Hence we may suppose, indif- 
ferently, 

Angle A OB = angle-arc ab, 
or Angle AOB = angle-arc ab -f- 360°, / 
or Angle A OB = angle-arc ab + 720°, / 

etc. etc. j ^ — j 2 — A 

If we put C for 360°, or the circum- \ 
ference, the general measure of the angle is ^ 

Angle AOB = iC -\- angle-arc ab, (1) " "" 

in which i may be any integer whatever, positive or negative. 

We may consider this same form to include the negative 
measure aMb. For, since ab -f- aMb = C, we have 

Arc ab = C — arc aMb. 
By substituting this value in (1) it becomes 

Angle AOB = (i + l)C- arc aMb. 




4 PLANE TRIGONOMETRY. 

Since we have i = — 2, — 1, 0, 1 ? 2, 3, etc., #<? infinitum, 

i-\-l may go through the same system of integral values as i. 

In general, if an angle is n° less than a circumference, we may 
call it, indifferently, 

Angle of (360 — n)° or angle of — n°. 

The general measure of the angle expressed in the form (1) 
has its most convenient application in Astronomy. The heavenly 
bodies perform unceasing revolutions, and thus describe continually 
increasing angles ; but each revolution brings them back to what 
we may consider the same position relative to the centre of motion. 

8. In order to give entire algebraic precision to the measure 

of an angle, we must suppose a distinction between the side from 

iVhich we measure and the side to which we measure. In all the 

preceding examples we have supposed the measure to be from OA 

to OB. Had we measured from OB to OA, the arc ab would 

have been described in the negative direction, or aM b would have 

been described in the positive direction. Hence we should have 

had 

Angle BOA = — arc ab or -(- arc aM b, 

which is the negative of the corresponding measure from OA to 
OB. Hence : 

By interchanging the sides we change the algebraic sign of 
the angle. 

To give uniformity to this mode of measurement, the side OA, 
from which we measure, is supposed fixed, while the other side 
varies in direction according to the magnitude of the angle. 

When angles are represented in a general way, the side OA 
may be conceived as extending out horizontally towards the right. 
Then the other side, OB, will have a definite direction for every 
angle we choose to assign. For example : 

For 90° the side OB will point upward. 
" 180° " " " " " to the left. 
" 270° " " " " " downward. 
« 360 o u u u « « to the right. 
« 450° " " " " " upward, 
etc. etc. 



OF GONIOMETRY. 



Counting the angles in the negative direction, 
For — 90° the side OB will point downward. 
« _ 180° " " " " " to the left. 
m _ 270° " " " " " upward. 
etc. etc. / 

9, Division into Quadrants. The circle [ 



which measures angles is, for convenience, \ 
supposed to be divided into quadrants, as in 
the figure. 

An angle between 0° and 90° is in the first quadrant. 
" " " 90° " 180° " " second " 

u « « 180 o u 270° « « third " 

« " « 270° " 360° " " fourth " 

Counting the angles negatively from OA, 
An angle between 0° and — 90° is in the fourth quadrant. 
" " " - 90° " - 180° " " third " 

« « « _ 180° " - 270° " " second " 

« " " - 270° u — 360° " " first « 



Exercises. 

1. From the point O emanate a set of 5 lines making equal 
angles with each other, and another set of 
6 lines making equal angles with each 
other, the line OA being common to the 
two sets. Compute the values in degrees 
of the ten angles A Ob, bOB, BOc, etc., d ~~~ 
to/OA. 

2. What is the value of that angle 
whose negative measure is numerically 
double its positive measure ? 

3. If a side starting from the zero point move through — 1905°, 
in what quadrant will it be found, and what will be the smallest 
positive measure of the angle % 

4. Two arms start together from the same position OA to turn 
round O, the one going in the positive direction, so as to revolve 




6 PLANE TRIGONOMETRY. 

in 60 seconds, the other in the negative direction, so as to re- 
volve in 36 seconds. At what angle and in what time will they 
meet? 

5. If two revolving arms start out together from the position 
0° in the same direction, the one going 5° a minute and the other 
8° a minute, through what arc will each have moved when they 
again come together? At what angle will they meet? If they 
continue turning, after how many revolutions of each will they be 
together at their starting point ? 

6. Four lines, a, b, c, d, emanate from the same point o, 
making angle boc = 2aob, cod = 2boc, doa = 2cod. What are 
the values of the four angles which they form ? 

7. If an angle of 140° is multiplied successively by — 2, — 3, 
— 4, — 5, — 6, — 7, in what quadrants will the respective multi- 
ples fall, and what will be the smallest positive measures of the 
several angles formed ? 

8. Show that the following pairs of angles are supplementary : 

90°+a?and 90° — x; 

270° -x and 270° -fa?; 

60° - x and 120° + x. 

9. Show that the following pairs are complementary : 

45° -a? and 45° + a?; 

225° -x and 225° + x; 

60° — x and 30° + x. 

The Division of Angles. 

10. Bisection. If the angle AOB = n°, and if Ob is its 

bisector, then A Ob = %n°. 

If the side OB revolves about O, and 
the side Ob also revolves in the same 
direction half as fast, then Ob will con- 
tinually bisect the angle AOB. 

When OB completes a revolution, re- 
turning to the position OB, the bisector / 
Ob will have moved through 180°, and c 
will therefore lie in the opposite direction, Oc. Another revolu- 




OF Q0S10METUY. 



tion of OB will bring the bisector to the position Ob again, yet 
another to Oc, and so on. Hence : 

Tlie general measure of an angle has two bisectors 180° apart. 






11. Trisection. If the side Ob is to continually measure one 
third the angle AOB as OB revolves, B 

then we must have \ ^ 

AOb = iAOB. 



If OB, starting from the position * 



/ 



in the figure, goes through one revo- 
lution, Ob will go through 120° to the 
position Oc. A second revolution of 
OB will bring Ob 120° farther, to Od, 
and a third to its first position, Ob, 
after which it will repeat its movements. Hence : 

One third the general measure of an angle has three special 
angular values differing by 120°. 

12. Division into n Parts. If, as OB revolves, Ob continually 
measures — of it, then every revolution of OB will turn Ob 

through - of a revolution. Hence : 

The nth part of the general measure of an angle has n special 
angular values. 

13. Analytic Deduction. It will be remarked that, in the 
preceding sections, what we take the wth part of is not the angle 
A OB, but the general measure of this angle. This will be clear 
from the following analytic deduction of the same result. 

Let the smallest measure of the angle A OB be a. Then the 
other measures of this angle (§ 7) will be 

a+C, a + 20, a + 3C . . . a + W. 
Dividing these quantities by n, the quotients will be 

I l+lc, °-+*-G, etc. (1) 

The (n + l)th angle of this series will be 

n ' n n l ' 



8 PLANE TRIGONOMETRY. 

a. 

which will correspond to the same position of OB as - does. 

The continuation of the series will be 

%+C + -C, ^+0 + -0, etc., 

showing that the positions will be continually repeated in regular 
order. 

Exercises. 

1. If AOB = 30°, or 30° + O, or 30°+ 20, etc., at what 
angles will \AOB fall? 

2. How many degrees between the minute-lines on a clock-face ? 

3. At what angles with XII. are the hour and minute hands of 
a clock together % 

4. If the hour-hand is so displaced that when the minute-hand 
is at XII. the hour-hand is 2 m past XII., at what angles will 
the hands be together % 

5. What values may two thirds of the general measure of an 
angle of 105° have ? 

6. If an angle is in the third quadrant, what are the limits be- 
tween which its bisectors must fall ? 

7. Between what three sets of limits must a be contained in 
order that 3a may fall in the fourth quadrant ? 

8. Show that while one sixth of the general measure of an 
angle has six different values, two sixths has only three values, and 
three sixths only two values. Show that this diminution arises 
from several values falling together when multiplied by 2 or 3. 
As an example, take the case when a = 48°; \a = 8°, 68°, etc. 

14. Natural Measure of Angles. The division of the circum- 
ference into 360° is entirely arbitrary, and any other angle than 
the degree may be taken as the unit. B 

In purely mathematical investigations, where / \ % 

no division into degrees is required, the length / \ 

of the radius is taken as the unit of measure. / \ 

This unit is called the radian. O^- 'x 

The radian is therefore the angle subtended ^^^^ch 1 " 3 ^^ 
by an arc whose length is equal to the radius. arc XB = radius ox 



OF QONIOMETBT. g 

To find the relation of this unit of angle to the degree, 
minute, and second, we note that the ratio of the entire circum- 
ference to the diameter is 3.141 592 65, etc. (Geom., Book 
VI., §5.) 

Hence its ratio to the radius is double this number, or 

6.283 1S5 3, etc. Since the circumference measures 360°, the unit 

360° 
radius will measure a OQO 1 >r Q , or 57°. 295 779 5 . . . . Hence 

v. Zoo lOO o 

1 radian = 57.295 779 5 ... . degrees. 

= 3437.746 77 ... . minutes. 

= 206 264.806 .... seconds. 

In mathematics we use the symbol 

circumference -i circumference 

n = —r. = j.- = 3.141 592 65 

diameter radius 

Hence, when we take the radian as the unit, 

\n represents an angle of 90° ; 

re " " " " 180°; 

2tt " « « « 360° = circumference ; 

2n7t " " " " n circumferences. 



Exercises. 

Considering the radius of the circle as unity, what is the length 
of circular arcs subtending the following angles ? 

1. 28° ir 15".6. Ans. 0.493 72. 

2. 14° 8 7 37".8 " 0.246 85. 

3. 22° 25' 53".4 " 0.391 51. 

4. 90° " 1.570 80. 

Note. In these exercises the angle is first to be reduced to a common 
denomination of measure, either degrees, minutes, or seconds. For instance, 
28° 17' 15".6 = 101835".6 = 1697'.26 = 28°.287 67. 

If the radius is 100 metres, how many degrees and minutes 
will arcs of the following lengths subtend % 

5. 100 metres. Ans. 57° Yi' 44".8. 

6. Y2 metres. " 41° 15'.1. 

7. 310 metres. " 177° 37'. 



10 PLANE TRIGONOMETRY. 

"With what radius will — 

8. Arc of 32 metres' length subtend an angle of 32° ? 

Ans. 57 m .296. 

9. Arc of 32 metres' length subtend an angle of 32' ? 

10. " " 32 " " " " " " 32" ? 

11. " « 1 metre " " " " " 67"? 

Ans. 3078 m .6. 

12. Two railways met at right angles at 0. They are to be 
connected by a quadrant PQ, of which the inner rail shall be 
600 metres in length. What is the common _ 
distance OP and OQ of the switches from the ' "^N J 
point in which the two inner rails would \ | 
meet? Ans. 381 m .97. \ 

13. In the preceding case, if the rails are 5 
feet apart, how much longer will the outer 
rail of the curve be than the inner one ? 

Ans. 7.854 feet. 

14. Show that if three circles, equal or unequal, mutually touch 
each other externally in the points A, P, and (7, the sum of the 
three included arcs AP -f- PC -f- CA, expressed in angular meas- 
ure, is equal to a certain constant. What is this constant ? 

15. If the two lesser circles, still touching each other, touch 
the greater one internally, show that the sum of their arcs minus 
the arc of the greater circle, expressed in angular measure, is equal 
to the same constant as that of the preceding problem. 

16. The earth's equatorial diameter being 12,756 kilometres, 
what is the length of one degree of the equator in kilometres and 
in miles, assuming 1 metre = 39.37 inches. 

17. Explain why a degree of latitude is greater at the poles 
than at the equator, although the radius of the earth is less. 

Bemark. — At this stage of his progress, if not sooner, the student should be 
familiarized with the use of the logarithmic and trigonometric tables, and should 
employ them in all computations in which they are applicable 



CHAPTEK II. 
THE TRIGONOMETRIC FUNCTIONS. 



The Sine, Tangent, and Secant. 

15. To investigate the numerical relations between the sides 
and angles of geometric figures, certain functions of angles are 
employed in trigonometry. These functions are defined in the 
following way : 

Let OX be that side of the angle from which we measure, the 
length OX being taken as the radius of a circle. Also, suppose 

OB, the side to which we measure ; 

M, the point in which OB intersects the circle ; 

XQ, the line tangent to the circle at X ; 

N, the point at which 
this tangent meets the 
side OB ; 

MP, the perpendicu- 
lar from M upon OX. 

Then taking the ra- 
dius OX as unity, and 
expressing other lengths \ 
in terms of this unit : i 
I. The length MP < 
is called the sine of the 
angle XOB. 

II. The length NX 
is called the tangent 
of the angle XOB. 

III. The length ON is called the secant of the angle XOB. 



/ 





12 PLANE TRIGONOMETRY. 

The absolute lengths of the lines representing the sine, tan- 
gent, and secant, considered as lines, will vary with the radius of 
the circle. This is avoided by taking for the sine, tangent, and 
secant, not the lines which represent them, but the ratios of 
the lines to the radius of the circle, which ratios will be pure 
numbers. 

"We have now to prove that these 
numbers are the same for the same 
angle whatever be the radius. 

Let XON' be the angle. 

From the vertex O draw the two 
arcs XM and X'M' with any two radii 
OX and OX. 

Erect the respective sines and tangents PM, XX, P'M', X'X'. 

Then because the triangles OPM, OXX, OP'M, and OXX' 
have the angle at common, and the respective angles at P, X, 
P', and X' all right angles, and therefore equal, these triangles 
are equiangular and similar. 

Comparing the sides about the equal angles we have the ratio 
PM : OM = P'M' : OM; \ 
XJST : OX = XN' : OX'-, \ (a) 

OX : OX = OX : OX'; ) 
Because OM = OX = radius of inner circle, 

and OM'— OX = radius of outer circle, 

we have, by definition, 

PM : OM = sine of POM; 
XX : OX = tangent of POM; 
OX : OX = secant of POM 

The equations (a) now show that the sine, tangent, and secant 
of the angle will be represented by the same numbers whether we 
measure them in the inner or outer circle. Therefore : 

To each angle of a definite magnitude corresponds 
One definite number, called the sine of the angle; 
Another definite number, called the tangent of the angle; 
Another definite number, called the secant of the angle. 



THE TRIGONOMETRIC FUNCTIONS. 13 

17. Notation. If we call a any angle, 

Its sine is written sin a ; 

" tangent " tan a ; 

" secant " sec a. 

Note. The representation of the trigonometric functions by lines is for the 
sake of clearness. They are not really lines, but ratios of lines which are pure 
numbers. But in studying these numbers the ideas are fixed by representing 
them by lines, as is done in some departments of algebra. We have only to 
remember that the lines are not the functions themselves, but lengths propor- 
tional to the functions, and therefore admitting of being used to represent the 
functions. These lengths are, however, really equal to the products of the radius 
by the corresponding functions. For example, if 

PM 
sin XOM = -fz: , 
Radius 

then sin XOM . Radius = PM. 

18. Eemarlc. The sine of an angle is equal to half the chord 
of twice the arc of the angle, the radius being supposed unity. 
Hence : 

Any chord in a circle is equal to the radius multiplied by 
twice the sine of half the angle subtended by the chord. 

Exercise. 

Let the student find by actual measurement with dividers and 
scale the sine, tangent, and secant of every 10° 
from 0° to 90° in the following way : 

With a radius equal to some unit or some 
whole number of units on a scale, describe the 
quadrant XB. Either 4 inches, 5 inches, or 
a decimetre would be a convenient radius. 

Divide the quadrant into 9 arcs of 10° 
each. Through each point of division, M 
for instance, draw a radius and continue it 
until it intersects the tangent at N. Then 
measure — 

1. The distance of each division-point on the arc from the line 
OX, which distance, divided by the radius, will give the sine of the 
corresponding angle. 



14 



PLANE TBIGONOMETRY. 



2. The distance of each point of intersection, JV, from X, which 
being divided by the radius will give the tangent of the angle. 

3. The length of each OX, which being divided by the radius 
will give the secant of the corresponding angle. 

The results should all be expressed in decimals to three places, 
exhibited in a little table in the following form, and afterward 
compared with the values found in the trigonometric tables : 



Angle. 


Sine. 


Tangent. 


Secant. 


0° 








10° 








20° 








30° 


. . . 






etc. 


■ • • 







With care the average deviation of the measures from the 
truth ought not to exceed .005, except in the cases of the tangent 
and secant of 70° and 80°, which are so great that they cannot be 
easily found in this way. 

19. To find, oy measurement, the angle corresponding to a 
given sine, tangent, or secant. 

Analysis. If a sine is given, the end of the arc correspond- 
ing to the required angle must be at a distance from the line OX 
equal to the given sine, the radius being unity. Y 
Therefore if we take on the perpendicular 
OY & distance OR equal to the product of 
the radius by the given sine, and through M 
draw a parallel to OX, the point in which it 
intersects the arc will give the required angle. 

To find the angle corresponding to a given 
tangent we take a distance XS equal to the product of the given 
tangent into the radius. Join OS. The angle XOS will be that 
required. 

For a secant we take the product of the radius by the secant in 
the dividers, and from O as a centre draw an arc cutting XN in 
a point N. Joining ON, the angle XON will be that required. 




THE TRIGONOMETRIC FUNCTIONS. 



15 



Exercises. 

1. Find by measurement the angles of which the sines are 
}, 0.3, 0.4, 0.6. 

2. Find arc-tan £, 1, 1.5, 3. 

The expression arc-tan is used for brevity to mean the arc corresponding to 
a given tangent. 

3. Find arc-secant 1.5, 2. 

20. Functions of unlimited angles. Thus far we have con- 
sidered only the sines, tangents, and 
secants of angles less than 90°; that is, 
of angles in the first quadrant (§9). 
As our angle increases to an entire cir- m jC 
cumference, the functions are deter- ' 
mined by the same construction modi- 
fied to suit the case. 

The following are the general defi- 
nitions: 

We first generalize the construction. 

On the sides of the angle we take equal lengths OX and OM 
as the unit of measurement and radius of the circle. 

At X we erect a line TXT' perpendicular to OX, extending 
indefinitely in both directions. 

We also suppose the revolving side OM to be produced indefi- 
nitely in both directions, and XO to be produced so as to form 
the diameter XTJ. 

Then, however the side OM may revolve — 

I. The sine of the angle XOM is always represented by the 
perpendicular from M upon the line OX. The sine is positive or 
negative according as M is above or below OX. 

II. The tangent of the angle XOM is always represented by 
the distance from X to the point in which the side OM produced 
intersects the line TXT' . The tangent is positive or negative 
according as the point of intersection is above or below X. 

III. The secant of XOM is the length of OM produced inter- 
cepted between O and the vertical line TXT'. The secant is 




16 



PLANE TRIGONOMETRY. 




positive or negative according as it is in the direction from to- 
ward M or in the opposite direction. The positive direction of 
OM is supposed to revolve with the side OM, and therefore to be 
always from O toward M. 

21. Changes in the value of the sine. If we suppose the 

side Om of the angle XOm to revolve around 0, carrying the 

sine mp with it, the latter will increase to 

its maximum value, equal to radius unity, 

when m reaches Y, and angle XOm = 

90°. Hence 

sin 90° = + 1. 

As m moves from Y to X' , the sine 
will diminish from 1 to zero. Because 
angle XOX = 180°, 

sin 180° = 0. 

If m passes over X into the third quadrant, the perpendicular 
M'P' will be below the line X' OX. This change of direction is 
expressed by changing the algebraic sign of the perpendicular from 
-f- to — . This is in accordance with the following general principle : 

Whenever distances measured in one direction are considered 
positive, those in the opposite direction are negative. 

Hence also : 

In the third quadrant the sine is negative. 

When the point m reaches the position Y' it will have 
moved through three quadrants or 270°, and the sine will coincide 
with the radius O Y ' of length unity. Hence 

sin 270° = — 1. (1) 

As m moves from Y' to X, the sine will increase from — 1 
to 0. Hence: 

I. In the fourth quadrant the sine is nega- 
tive. 

II. sin 360° = sin 0° =- 0. 

The changes of algebraic sign as the angle 
goes through the four quadrants are shown in the 
annexed diagram. 

Angles having equal sines. If angle X ' OM = XOm, the 




Algebraic signs 
of the sine. 



THE TRIGONOMETRIC FUNCTIONS. 17 

two angles XOm and XOM will be supplementary. Also in this 

case the triangles XOm and X' OM will be identically equal; so 

that 

PM — pm. 

Now PM represents by construction the sine of XOM, and 

pm the sine of XOm. Hence : 

The sines of supplementary angles are equal. 

In symbolic language this theorem is expressed thus : If a be 

any angle, then 

sin (180° — a) = sin a (2) 

and sin (90° + a) = sin (90° — a). 

If the points M ' and M" are equally distant from Y' , so that 

angle M'OY' = angle Y'OM", which angle call y, the sines 

P'M' and P"M" will be equal. Hence, whatever be y, 

sin (270° -y) = sin (270° + y). 

22. Changes in the tangent. As the line Om revolves round 
O and m approaches Y, the point of intersec- 
tion JV will move upwards without limit. As 
m reaches Y, Om will become parallel to the 
tangent line, and the point H will recede to ^/^ 
infinity. Hence : / \ 

The tangent of 90° is infinite. ~x\ 

When m is in the second quadrant, suppose V 
in the position 31, the revolving side OM will \^ 
not intersect the tangent line at all in the posi- Y '" 

tive direction OM. We must therefore suppose the revolving line 
to be produced in the negative direction ON' so as to intersect 
the tangent line at N' below X. The distance XN' is then to be 
regarded as negative. Hence : 

In the second quadrant the tangent is negative. 

Following the motion, we see that when m reaches X', N' 
reaches X and the tangent becomes zero. Hence 

tan 180° = 0. 
When m is in the third quadrant, impasses above X and the tan- 
gent is positive, so that 

In the third quadrant the tangent is positive. 



18 PLANE TRIGONOMETRY. 

Continuing the reasoning, we see that the tangent of 270° is 
infinite, and that in the fourth quadrant the tangent is negative. 

23 Changes in the secant. The secant is defined as the 
distance ON from to the point N, in which the revolving side 
intersects the tangent line XN. 

When m falls on X and the angle is zero, the secant is equal to 

OX, or unity. Hence 

sec 0° = 1. (3) 

As m moves from Xto Y, the secant increases without limit 
and becomes infinite when m reaches Y. Hence 

sec 90° = oo . (4) 

As m moves from Y through X' to Y\ the intersection of 
the revolving line with the tangent line falls in the negative part 
of OM, or in the direction ON' . Hence : 

In the second and third quadrants the secant is negative. 

At Y', when the angle is 270°, the secant again becomes in- 
finite. 

Between Y f and X, or in the fourth quadrant, it is again 
positive. 

24. If we suppose the revolving line to make an integral 
number of revolutions from any point, it will return to its original 
position, and all the trigonometric functions will have the values 
corresponding to that position. Hence, if O is a circumference 
and n any integer, 

sin (nC -f- a) = sin a ; 1 

tan (nC + a) = tanar ; \ (5) 

sec (nO -\- a) = sec a. J 
In other words, 

The values of the -trigonometric functions wre not altered ~by 
increasing the angle by any integral number of circumferences. 

If the angle is increased indefinitely, the values of these func- 
tions continually repeat themselves. This fact is expressed by 
saying that these functions are periodic. 



II IK TRIGONOMETRIC FUNCTIONS. 19 

Exercises. 

Prove the following expressions for the trigonometrical func- 
tions of angles of more than 00° by the necessary diagrams. The 
angle x may be supposed less than 90°, though this restriction is 
not necessary to the validity of the formulae. 

1. sin (90° + x) = sin (90° - a?); 1 

2. sin (180° + x) = - sin x ; [ (6) 

3. sin (270° + x) = — sin (90 c - x). J 

1. tan (90° + x) = - (tan 90° - x)\ 1 

5. tan (1S0° + x) = tan x ; [ (7) 

6. tan (270° + x) = - (tan 270°- a?). J 

7. sec (90° + a?) = — sec (90°— x); 1 

8. sec (180° + x) = — sec x; \ (8) 

9. sec (270° + x) = — sec (270°— a?). J 

10. sin (—a?) — — sin a?; 1 

11. tan (— a?) = — tan x\ \ (9) 

12. sec (— x) = sec x. J 

Note. When we have the values of the trigonometric functions from to 
90°, we can by these formula? find the values for all angles. 

The Cosine, Cotangent, and Cosecant. 

25. In the preceding sections we have supposed the side of 
the angle from which we count the degrees to go out toward the 
right, and the positive direction of motion to be opposite to that 
of the hands of a watch. But this restriction is only to fix the 
thought. We may suppose the angle to have any situation and 
to be counted in any direction without changing the values of the 
sine, tangent, and secant, provided that we reckon the lengths of 
the lines representing the functions in the right way. 

Let us count the angle from O Y in the direction toward OX. 
The tangent line must then touch the circle at Y, and its positive 
direction must be toward the right. 

Then (radius O Y = 1) the lines 

PM= sin POM, ] 

Ylf= tan POM, \ (a) 

ON = sec POM, J 




20 PLANE TRIGONOMETRY. 

will have the same values as in an angle equal to POM counted 
in the usual way from OX toward Y. 

Moreover, the changes of sign will 
he the same as before through the whole 
circle, namely : 

From X to Y' (now the second 
quadrant) the sine PM will be positive 
because it is measured to the right. 

From Y r to X' (the third quadrant) 
it will be negative because it is measured toward the left. 

It will also be negative from X' to Y (now the fourth quadrant). 

The corresponding propositions can be shown for the tangent 
and secant. 

Now because angle XOM+ angle MO Y= angle XOY— 90°, 
MO Y is the complement of XOM. Therefore the equations (a) 
may be written 

PM = sin comp. of XOM; 

YW— tan comp. of XOM; 
OX = sec comp. of XOM. 

Because when we have an angle its complement can always 
he determined by subtracting it from 90°, we can always find the 
sine, tangent, and secant of the complement when we know the 
angle. Therefore the sine, tangent, and secant of the comple- 
ment of an angle may be regarded as three additional trigono- 
metrical functions of the angle itself. They are named thus : 

The sine of the complement is called the cosine of the angle. 

The tangent of the complement is called the cotangent of 
the angle. 

The secant of the complement is called the cosecant of the 
angle. 

Thus the new functions are defined in the following way : • 
cosine a = sin (90° — a); 1 
cotang a = tan (90° — a); I (10) 

cosecant a = sec (90° — a). J 

The words cosine, cotangent, and cosecant are abbreviated to 
cos, cot, cosec, respectively. 



THE TRIGONOMETRIC FUNCTIONS. 



21 



The forms (10) enable us to find the cosine, cotangent, and 
cosecant of an angle when we know the sine, tangent, and secant 
of its complement. Thus if the cosine of 60° is required, we have 
cos 60° = sin (90° - 60°) = sin 30°. 
Also, by substituting 90° — a for a, we find 
sin a = cos (90° — a); 1 
tan a = cot (90° - a); I (11) 

sec a = cosec (90° — a). J 
The versed-sine and co-versed-sine. Besides these six func- 
tions, two others, the versed-sine and co-versed-sine, are sometimes 
used. Their definitions are : 

Versed-sine = PX = 1 — cosine ; 
Co- versed-sine = O Y — PM = 1 — sine. 

26. The six trigonometrical functions may be represented on 
a single diagram. The 

functions as written are X cotajw^nt^y^ 

all those of the angle 
XOM. For the secant 
and cosecant we have 

ON '= see XOM, 
ON' = cosec XOM, 
because XOM is the 
complement of MO Y. 

Because PM || OP' 
and OP || P'M, there- 
fore OP = P'M, so that 
we may take either OP or P'M as the cosine of XOM. 

27. The general definitions of §20 may be extended to com- 
plementary functions, thus : 

Having OY _L <9Xand OY= OX 
= 1, we draw through Y an indefinite 
line YT parallel to OX. Then— 

IV. The cosine of any angle XOM 
is represented oy the distance OP to 
the foot of the sine, positive or negative 
according to its direction. 





22 



PLANE TRIGONOMETRY. 



Y. The cotangent of any angle XOM is represented by the 
length from Y to the point in which OM produced inter- 
sects YT. 

VI. The cosecant of XOM is represented by the length of the 
side OM, intercepted between O and the line YT. 

The algebraic signs of the several functions in the four quad- 
rants are shown in the following diagram. 

Tangent. 



Sine. 



Secant. 



' + -f- 



/ X 1 

— + > 

+ — J 



2// 


^Ni 


/ — 


+ \ 


V — 


+ / 


3 \__ 


^/ 4 



Cosecant. Cotangent. Cosine. 

28. The following are the limiting values of the trigonometri- 
cal functions : 

I. Bine and cosine. The sine MP and cosine OP are neces- 
sarily not greater in absolute value than OM= 1. The limits of 
these functions are therefore -f- 1 and — 1. 

II. /Secant and cosecant. Since the tangent line lies without 
the circle, a secant can never be less than unity in absolute magni- 
tude. But we have found that it may increase to infinity in either 
the positive or negative direction. Hence the limits of the secant 
and cosecant are 1 and infinity, and — 1 and — oo . 

III. Tangent and cotangent. The limits of the tangent are 
easily seen to be — oo and -J- oo , or a tangent and cotangent may 
have any value whatever. 

29. When we know the numerical values of the sine, tangent, 
and secant of all angles from 0° to 90°, we have the values of all 
six functions of any angle whatever, because as we go around the 
circle the values of the functions are simply repetitions of the 
values between 0° and 90°. 

Let a be any angle less than 90°. Then any angle in the first 
quadrant may be represented by a. 

In the second quadrant it may be represented by 180° — a. 

" third " " " " 180° -for. 

" fourth " " " " 360° -a. 



THE TRIGONOMETRIC FUNCTIONS. 



23 



The student should now have no difficulty in demonstrating 
the following relations by completing the 
construction indicated in the margin and 
attending to the general definitions §§ 20, 
27. Some of these relations have already 
been given or explained. 

Second Quadrant 



(12) 




sin (180° - a) = 


sin « ; 


cos (180° — a) = 


— cos a: = — sin (90° — a); 


tan (180° — a) = 


— tan a ; 


cot (180° — *) = 


— cot a = — tan (90°— a); 


sec (180° — a) = 


— sec a ; 


cosec (180° — a) = 


cosec a = sec (90° — <*). , 



Third Quadrant. 

sin (180° + a) = — sin a ; 
cos (180° + a) = — cos a = — sin (90° — a); 
tan (180° + a)= tan ^ ; 
cot (180° + a) = cot a = tan (90° - or); 
sec (180° -(- or) = — sec a ; 
cosec (180° -\- a) — — cosec « = — sec (90° — a) 

Fourth Quadrant. 

sin (360° — a) = — sin a ; 
cos (360° — a) = cos « = sin (90° — a); 
tan (360° — a) = — tan a ; 
cot (360° — a) = — cot a = — tan (90°— a); 
sec (360° — a) = sec a ; 
cosec (360° — a) == — - cosec a = — sec (90°— a). . 



(13) 



(14) 



We may equally express the six functions of all angles in terms 
of the six functions of angles not greater than 45°. Let y repre- 
sent any angle not greater than 45°. We may then represent 
Any angle from 0° to 45° by y ; 

« " 45° " 90° " 90° — y\ 



90 c 



135 c 



90° + y; 



24 



PLANE TRIGONOMETRY. 



Any angle from 135° to 180° by 180° — y; 
« « 180 o u 225° « 180° + y 5 
" " 225° " 270° " 270° - y ; 
« « 270° " 315° " 270° + /; 
" " 315° " 360° " 360° — y. 
Then, in addition to the relations (12), (13), and (14), which will 
remain true when we write y instead of <*, we shall have the fol- 
lowing, which the student should prove. 

To do this let the student suppose that in the diagram §26 angle XOM=y, 
and let him construct the six functions for angles of 90° + ^, 270° - y, etc., 
and compare the lines representing them with the lines on the diagram of § 26. 
The set corresponding to the first quadrant are already given in (10) and (11). 



Second Quadrant. 
sin (90° -\- y) =z cos y ; 
cos (90° -\- y) = — sin y ; 
tan (90°+.y;j } = — cot y; 
cot (90° + y) — — tan y ; 
sec (90° -f- y) = — cosec y ; 
cosec (90° + y) = sec y. 



(15) 



Third Quadrant. 

sin (270° — y) = — cos y ; 
cos (270° — y) = — sin y ; 
tan (270° — y) = cot y ; 
cot (270° — y)= tan/; 
sec (270° — y) = — cosec y ; 
cosec (270° — y) = — sec y. 

Fourth Quadrant. 

sin (270° + y) = — cos y ; 
cos (270° + y)= siny; 
tan (270° + y) — ■— cot y ; 
cot (270° + y) = — tan ^ ; 
sec (270° -(- ^) = cosec y ; 
cosec (270° -\- y) = — sec /. 



(16) 



(17) 



THE TRIGONOMETRIC FUNCTIONS. 



25 



Among the preceding forms of this chapter, the following are 
of especially frequent application : 

sin a = cos (90° — a) = cos (a — 90°) = — cos {a -f 90°); ] 
cos a = sin (90° - a) = - sin (a - 90°) = + sin (a +90°). J ( 18 ) 

Exercises. 

1. Express the six functions of the following angles in terms of 
the three functions sine, tangent, and secant of angles less than 90° : 



sin 105° 


; 200° 


; 295°. 


cos 105° 


200° . 


295°. 


tan 105° 


; 200° 


295°. 


cot 105° 


200°. 


295°. 


sec 105° ; 


200°: 


295°. 


cosec 105° 


; 200° 


; 295°. 



2. The following table shows the values of four of the func- 
tions for every 10° of the first 40° to two places of decimals. By 
means of these values extend the table to 360°, showing the values 
of all four functions for each angle : 



Angle. 


Sin. 


Tan. 


Cot. 


Cos. 


0° 
10° 
20° 
30° 
40° 
50° 
60 c 
70° 
etc., to 
360° 


0-00 

+ 0-17 
+ 0-34 
+ 0-50 
+ 0-64 


0-00 
+ 0-18 
+ 0-36 

+ 0-58 
+ 0-84 


00 

+ 5-67 
+ 2-75 
+ 1-73 
+ 1-19 


+ 1-00 
+ 0-98 
+ 0-94 
+ 0-87 
+ 0-77 



3. Demonstrate the relations (18) by drawing a diagram show- 
ing an arbitrary angle a and an angle 90° greater and less, with 
the lines representing the sines and cosines. 

4. What relations subsist between the following pairs of func- 
tions ? 

(a) sin (45° -|- a) and cos (45° — a) ; 

(b) sin (135° + a) and cos (135° - a) ; 

(c) tan (225° + a) and cot (225° - a) ; 
(i) sec (315° + a) and cosec (315° — a). 



26 



PLANE TRIGONOMETRY. 



30. Special values of trigonometric functions. If angle 
XOM = 45°, we shall also have OMP = 45°, and therefore 

OP 2 + PM 2 =± 2 PM 2 = 0M\ ^ 

whence PM= 0MV\. 

Therefore sin XOM = sin 45° = V\. (a) 

In the same way we find 

XT = OX, 
whence tan XOM = tan 45° = 1. (b) 

Again, OT 

whence sec XOM 




VOX' + XT 2 = V2 OX; 
sec 45°= V2. 
'Next, let angle XOM = 30°. 
Make angle XOM ' = XOM = SO . 
The triangles M'OM' and TOT' then have 
each of their angles 60°, and so are equilateral. 
Therefore MP = i MM' = iOM = iOX. 
Hence sin 30° = f (d) 

In the same way 

XT = i 0T 9 
OT' — XT 2 = OX 2 = 1; 
2 



(') 




\0M V 



or 


$0T 2 = 1, 0T=Vi = ^ 


xr= 


1/3' 






4/3 
2"' 




Also, 


OP = VOP 2 - MP 2 = VI - i = 




Hence 


tan 30° = —zz, 

Vs 




w 




2 
sec 30° = —p., 

Vs 




00 




Vs 

cos 30° = -£-. 




w 



Functions of '18°. It is shown in geometry that if the radius 
of a circle be divided in extreme and mean ratio, the greater seg- 
ment will be the chord of 36° ; that is, twice the sine of 18°. 

Putting 1 for the radius and r for the greater segment, the 
condition that the division shall be in extreme and mean ratio is 

1 : r :: r : 1 — r, 



THE TRIGONOMETRIC FUNCTIONS. 27 

or, equating the product of the means to that of the extremes, 

? >2 = 1 — r. 
The solution of this quadratic equation gives 

-\±Vb 
r= g . 

The positive root is the only one we want. Hence 

4/5-1 



sin 18° = 



We then find 
cos 



18° = radius 2 - sin 2 18° = 1 - sin 2 18°. 



Hence cos ^ 



= VlO+2^ 



JrJ 



31. Angles corresponding to given trigono- 
metric functions. When the value of a trigonometric func- 
tion is given and the angle is required, there are always two 
solutions to the problem. 

The Sine. It has already been shown that two supplementary 
angles have the same sine. Hence if a is an angle corresponding 
to a given sine, 180° — a will be another angle equally correspond- 
ing to it. 

We may also say that if 90° — /3 be an angle corresponding to 
any sine, 90° -f- fi W1U a l so correspond to it, because the sum of 
these two angles is 180°. The same statement applies to the 
angles 270° - fi and 270° + p. 

Unless there is some restriction upon the angle to be chosen, 
we cannot decide which angle to take. The most common restric- 
tion is that the angle must be between the limits — 90° and + 90°, 
or must be in either the first or fourth quadrant. There will then 
be between these limits one angle and only one for a given sine. 
Since the measurements of latitude on the surface of the earth are 
restricted between limits — 90° and -f- 90°, the latitude of a place 
is completely fixed by its sine. 

The Cosine. The construction of the cosine shows that it has 
equal values for positive and negative angles. Hence if a be an 
angle corresponding to a given cosine,— a or 360°— a will equally 
correspond to it. 



28 PLANE TRIGONOMETRY. 

Hence when an angle is determined by its cosine, either of 
these two angles may be chosen unless some restriction is placed 
upon the choice. The most common restriction is that the angle 
shall be positive and less than 180° ; that is, in the first or second 
quadrant. For every cosine there will be one and only one angle 
between the limits 0° and 180°. 

The Tangent. Since two angles which differ by 180° have the 
same tangent, it follows that if a be an angle corresponding to any 
tangent, 180° -|- a will equally correspond to it. Hence there is 
always a choice between these two angles, unless some restriction 
is placed upon the angle. 

The Cotangent. The cotangent being determined, like the 
tangent, by the intersection of the revolving side with a tangent 
line, every pair of angles corresponding to the same tangent will 
also correspond to the same cotangent. Hence a and 180° -f- & 
always have the same cotangent. 

The Secant. We readily see that the angles a and — or or 
360° — a have the same secant. Hence if a: be an angle corre- 
sponding to a given secant, 360° — a will be another angle corre- 
sponding to that same secant. 

The Cosecant. From the diagram (§ 27) it is easy to show 
that any two supplementary angles have the same cosecant. 

Exercises. 

What other angles have the same sines as the following ? 

1. 105° ; 2. 185° ; 3. 290°. 
What other angles have the same cosines as the following ? 

4. 72°; 5. 165°; 6. 320°. 

What other angles have the same tangents as the following % 

7. 50°; 8. 205°; 9. 355°. 

10. A pair of angles having the same sine differ by 24°. What 
angles are they ? 

11. A pair of angles having the same cosine differ by 110°. 
What angles are they ? 

Note. There are two pairs of angles which answer each of the two last 
questions. 

Find two values of a from each of the following equations : 



TEE TRIGONOMETRIC FUNCTIONS. 29 

12. sin a = — cos 23°. Ans. a = 247° or 293°. 

13. cos a — — sin 23°. 

14. sin a = cos ft. Ans. a = 90° — ft or 90° + ft. 

15. tan a = cot 72°. 

16. cot or = — tan 175°. 

17. tan a = — cot /3. 

18. sec a = cosec 32°. 

19. cosec** = — sec 32°. 

20. sec a = cosec ft. 

21. sin a = sin 23°. 

22. sin a = cos (90° — x). Ans. or = x or 180° — a?. 

23. sin a = cos (90° + x). Ans. a: = — x or 180° -f x. 

24. sin <* = cos (270° + x). 

25. sin a = cos (270° — x). 

26. cos a = cos (180° — x). 

27. costf= cos (180° + a). 

32. Extension to unlimited angles. Since when any inte- 
gral number of circumferences is added to an angle its trigono- 
metric functions remain unaltered, we must, to find the most 
general expression for the angle corresponding to a given function, 
add an arbitrary number of circumferences to the angles found in 
the last section. Then the most general expression for angles 
which have the same sine will be 

7i<7+90° + or and nC+ 90° — a, 
in which n may take all integral values, positive and negative, 
including zero, and a must have such a value that 90 — a shall 
correspond to the given sine. 

This statement also means that all the angles formed by giving 
different values to n, while a remains constant, will have the 
same sine. 

Example. Let us suppose the given sine to be that of 65°. 
Then a = 25°, and the pairs of angles 

65°, 115°; - C+65°, - (7+115°; 

(7+65°, 0+115°; —2(7+65°, -2(7+115°; 

2(7+65°, 2(7+115°; etc. etc., 

etc. etc.; 

will all have this same sine. 



30 PLANE TRIGONOMETRY. 

Exercises. 

1. Having given 

sin 2a? = J, 

find the four corresponding values of x within the first circumfer- 
ence. Note that sin 30° = J. Ans. 15°, 75°, 195°, 255°. 

2. Having given 

sin 2a == VJ , 
find the four corresponding values of a. 

First find the two values of 2a, and then, by §§ 10-13, the two values of half 
of each of these angles. 

3. Having given 



cos 2/3 



find the four corresponding values of /3. 
4. Having given 



cos 2/3 



find the four corresponding values of /3. 

5. If tan 2a — — 4/3, find four values of a. 

6. Show that if the value of tan -| a be given there will be only- 
one value of a to correspond to it within the first circumference. 

7. If sin \x = sin 15°, 

find what two values may x have, and show that these values will 
have the same cosine. 

8. Form the most general expression for all angles having the 
same cosine. 

9. Form the most general expression for all angles having the 
same tangent. 

Relations bet-ween the Six Trigonometric Functions. 

33. When we know any one trigonometric function of an 
angle two definite values of the angle can then be determined, by 
constructions like those of § 19. Knowing the angle, the values 
of the other functions can be found. Hence from any one func- 
tion of an angle all the others can be found. We have now to 
investigate the algebraic relations by which this may be done, 
seeking to express each function in terms of the five others. 

Let us put 

a == angle XOM. 



THE TRIGONOMETRIC FUNCTIONS. 



31 



Because OPM is a right-angled triangle, 
OP' + PM 2 ^ 0M\ 
or 



Cofasi& jWiuV / 



\OMl^ \OMJ 
Hence 

cos a a -\- sin 3 a = 1, 
which gives 




cos a = ± Vl — sin 2 a. (1) 

The significance of 

the double sign ± is this : 

whenever a sine is given 

there will be two angles to correspond to it, of which one will be 

as much less than 90° as the other is greater. The cosines of these 

angles will be equal with opposite algebraic signs. 

The similar triangles OPM and OXN give 

OP:PM=OX:XN; 

OP: OM = OX: ON. 

Substituting for the lines their algebraic equivalents, the first 

proportion gives 

cos a : sin a = 1 : tan a. 

sin a sin a 

(2) 



Hence 



tan a 



cos a 



VI- 



sin oc 



which gives the tangent in terms of the sine and cosine. 
The second proportion gives 

cos a : 1 = 1 : sec a, 
1 1 



whence 



sec a = 



vl — sin 2 a 



(3) 



cos a 

"We conclude from this : 

The product of the cosine cmd secant is equal to unity. 

In other words, the secant and cosine are reciprocals of each 
other. 

By a similar cou-rse of reasoning upon the complementary tri- 
angle we find that the cosecant and sine are reciprocals of each 
other. 



32 PLANE TRIGONOMETRY. 

The similar triangles OP 'M and YN' give 
OP' :P'M = OY: YN' \ 
but OP' = PM=sma, 

whence sin a : cos a = 1 : cot a, 

cos or 4/1 — sin 8 a 
and cot a = -. = : . (4) 

sin a sin a v J 

Comparing with (2) we have the relation : The product of the 
tangent and cotangent of any angle is unity. 

In other words, the tangent and cotangent are reciprocals of 
each other. 

We thus reach the general conclusion that the three comple- 
mentary functions are each the reciprocal of one of the three other 
functions, namely : 

cosine = -: 

secant ' 

1 

cotangent = -7 7 ; 

tangent ' 

sine' 

Exercises. 

1. If sin y = 0.60, find cos y, tan y, cot y, sec y, and cosec y 

2. Find the values of the same functions when cos y = 0.60. 

3. Prove that the mean proportional between a cos x and 
a sec x is a. 

4c. Prove that the mean proportional between a tan ft and 
h cot fi is Vab. 

34. Expression of each function in terms of the others. By 
means of the relations (1) to (4) any one trigonometric function 
may be expressed in terms of any other by algebraic substitutions. 

The following are the expressions which we thus obtain. All 
not already given should be deduced by the student as an exercise. 

sin a - 



cosecant = 





a 


I 












_L - — LOb 


VI + tan 2 


a 


= 


VI + cot 2 
1 


a 




Vsec 2 a — 
sec a 


1 






cosec a 





THE TRIGONOMETRIC FUNCTIONS. 33 

1 cot a 



cos a = Vl — sin 2 a = 



Vl + tan 2 a Vl + cot* a 
1 Vcosec 2 a — 1 



tan <x = 



sec <z cosec or 

sin a Vl — cos 2 a 1 



y 1 — sin 2 a cos « cot a 

1 



= y sec 2 or — 1 = 



l^cosec 2 oc — 1 

|/1 — sin 2 « _ cos a: __ 1 

COt " 7" sln~^ Vl - cos 2 a tana 

1 , 

— / = == — y^cosec 2 a — 1. 
y sec 2 a — 1 

1 1 



sec « = = == = Vl 4- tan 3 a 

1/1 — sin 2 a cos or 

y 7 ! + cot 2 ^ cosec a 



cosec a 



cot a Vcosec 2 a — 1 

1 1 Vl + tan 2 ar 



sin or Vl — cos 2 or tan <* 

,/t-i it- sec « 

= y 1 + cot 2 a = =• . 

^ Vsec 2 or - 1 

Note. In algebraic work of this sort the student will find it convenient, 
instead of writing sin a, cos a, etc., in full, to use a single symbol for each 
function; for example, he may put: 

s = sin a ; — = cosec a. 
s 

t = tan a ; — = cot a. 

1 

c = cosa; — — sec <x. 
c 

It will be seen that the second members of most of these equa- 
tions are surds, showing that their values may be either positive 
or negative. The result may be expressed thus : 

Whenever one trigonometric function is given, the four other 
functions which are not its reciprocal may have either of two 
equal values with contrary signs. 



34 



PLANE TRIGONOMETRY. 



This arises from the fact that every such function may belong 
to either of two angles, and affords an example of the correspon- 
dence between geometric and algebraic results. 



Exercises. 

1. From the special values of sin 30° and sin 45° found in 

§ 30, namely, 

sin 30° = i, 

sin 45° = VJ, 
find the values of the other five functions of 30° and of 45°. 

Ans. cos 30° == — ? ; cos 45° = VJ. 

tan 30°= -V; tan 45° = 1. 

1/3' 

cot 30°= VS; cot 45° = 1. 



sec 30° 



sec 45°= V2. 



_2_ 

cosec 30° = 2 ; cosec 45° = V2. 

2. It has been shown that the three products sin X cosec y 
tan X cot, and cos X sec are each unity, -y N" 
When we replace the functions by the 
lines which represent them, the pro- 
ducts represent the areas of the rect- 
angles contained by the lines, and unity 
is replaced by the square of the radius. 
Hence we have the following theorems, 
which are to be proved by the similar triangles in the accompany- 
ing figure, where the construction is that of the trigonometric 
functions. 

I. Eectangle XT. YJV= OX% corresp. to tan X cot = 1. 

II. Eectangle MP.ON = OX\ " " sin X cosec = 1, 

III. Eectangle OP.OT ^ OX\ " " cos X sec = 1. 

3. From the value of sin 18° in § 30 find the sides of the regu- 
lar inscribed and circumscribed decagons of a circle of radius a. 




TUB TRIGONOMETRIC FUNCTIONS. 35 

Prove the following relations : 

4 i i « cos 3 a? 

4. 1 + sin x = ; — . 

1 — sina? 

sec -f- cosec 6 1 -\- cot tan 0-\-l 

sec — cosec ~ 1 — cot 6 ~ tan 0—1' 

sec 2 6 -\- cosec 2 



tan d + cot = 



sec cosec 



7. sec 6 -\- tan = s — ^. 

sec — tan u 

tan + sec — 1 

8. : 7T- 1 tt-t-z = tan 6 4- sec 0. 



9. 



/a zr~r*T = tan + sec 

tan 6 — sec + 1 ' 

tan — sin sec 



sin 3 6 ~ 1 + cos 6' 

10. "What angle is that of which the tangent is double the sine ? 

11. What must be the value of the cosine in order that the tan- 
gent may be n times the sine I Ans. -. 

12. Prove (r cos xf -f- ( r sin x sin uf + (^ sin a? cos w) 2 = ?* 2 . 

13. Prove {a sin ;/) 3 + (# cos y sin #) 2 -f- (a cos y cos 6) 2 = a\ 

14. Prove (cos a cos J — sin a sin 5) 2 

+ (sin a cos 5 4" cos a sm V — 1- 

15. Of what angle is the secant double the sine ? 

16. Of what angle is the secant four times the cosine? 



CHAPTER III. 
OF RIGHT TRIANGLES. 



35. Fundamental relations. Let ON be a right triangle of 
which a and o are the sides which contain the right angle, c the 
hypothenuse, a and fi the angles opposite a and b respectively. 

N 





If we take ON as a radius and draw the arc iOT from the 
centre 0, the side NO will, by definition, represent the sine of 
XON, and 00 its cosine, when the radius is ON. That is, 

NO .00 _ 

"We may show in the same way, by taking N as the centre and 
NO as the radius. 



00 



sin /?; 



NO 



cos /?. 



(2) 



We might also have deduced these equations from (1), because 
/3 = 90° — a, whence 

sin /? = sin (90° — a) = cos or ; 
cos J3 = cos (90° — a) — sin or. 
Again, by taking 00 as a radius, we find 



0(7 



= tan a = cot /? ; 



(91 



= tan j3 = cot or. 



(3) 



OF MIGHT TRIANGLES. 



37 



Putting NC = a, 00 = b, ON = c, the equations (1), (2), 
and (3) give the relations 

a = c sin a =5 tan a\ 



(4) 



b = c cos a = a cot <*; 
c = a cosec a = b sec or. _ 
We may express the same relations in terms of /?, using the 
complementary functions, as follows : 

a = c cos J3 = b cot fi ; -i 
J = c sin f3 = a tan j3 ; L (5) 

c = a sec /? = 5 cosec /?. J 
These relations may be summed up in the following general 
theorems : 

I. The hypothenuse of any right triangle is equal to a side into 
the secant of its adjacent angle or the cosecant of its opposite angle. 
II. A side is equal to the hypothenuse into the sine of the 
opposite angle or the cosine of the adjacent angle. 

III. One side is equal to the other side into the tangent of the 
angle adjacent to that other side or the cotangent of the angle ad- 
jacent to itself 

Exercise. 

Show by the above equations how each side will be expressed 
in terms of the others when a == 30° and when a = 45°, using the 
values of sin a, etc., already found — namely, 

sin 45° = 1/J; tan 45° = 1 ; etc., 

and sin 30° = -§ ; cos 30° = —- ; 

and show how the results agree with those of elementary geometry. 

36. Examples and exercises in expressio?i. In the accom- 
panying figure OQN, ONP, and OXN Q 
are right angles, and we put 
a = angle XON\ 
fi = angle NOQ. 
It will also be noticed that 
Angle XNP = a 



tan 30°=—^; etc.— 



and Angle ONX = XPJST = 90° - a. 




38 PLANE TRIGONOMETRY, 

It is now required to express all the other lines in terms of 
OX and trigonometric functions of a and /3. 
Solution. We have 

OX=OXsecor; 

XN = OX tan or; 

OP = OX sec a = OX sec 2 a; 

NP = OP sin a = OX sin a sec 2 a = OX tan « sec or; 
or XP = OX tan a = OX sec a: tan a (as before) ; 

OQ = OX cos /? = OX sec « cos /? ; 

NQ = OX sin p = OX sec a sin /?. 

Exercises. 

1. By the same process express 0$, QN~> OX, XX, XP, and 

XP in terms of OX and trigonometric functions. 

2. Express the same quantities in terms of XP. 

3. Express XX separately in terms of OX and XP, and by 
multiplying the two values prove the geometric theorem that 
XX is a mean proportional between OX and XP. 

4. In a right triangle the sides which contain the right angle 
are a and b, (a > b), and d is the difference of the angles at the base 
(hypothenuse). Express the length of the perpendicular from the 
vertex upon the base in two ways, and the lengths of the segments 
into which it divides the base each in one way. The expressions 
are all to be in terms of a, b and o\ 

Ans. (in part). One expression for the perpendicular is 
2? = a sin (45° — 6). 

Solution of Right Triangles. 

Since in a right triangle one angle — the right angle — is given, 
only two other independent parts are required t© solve the tri- 
angle. These two parts may be any two of the sides or one side 
and one angle. What parts soever are given, the remaining parts 
may be found by the equations (4) and (5). The following are all 
the essentially different cases. 



OF RIGHT TRIANGLES. 39 

37. Case I. Given the two sides a and b adjacent to the 

right angle* 

The first equation (4), 

a = b tan a, 

a 
gives tan a = r- 

Therefore the quotient of the two sides gives the tangent of 
the angle opposite the dividend side. 

From the tangent the angle a is itself found by the trigo- 
nometric tables; then sec a. or cos a; then the hypothenuse c 
from the equation 

c = b sec a 



cos a 

Example. Given a = 9 metres, b = 12 metres, to find the 
remaining parts of the triangle. 

Solution by numbers and measurement. 
tan a = T \ = 0.75. 

On the tangent line XN ( § 22) measure a distance from X 
equal to 0.75 of the radius OX; join the end of the distance to 0, 
and measure the angle XON which the joining line makes with 
OX. This angle will be a. The length of the joining line divided 
by OX will be the secant, which multiplied by b ■•= 12 will give 
the hypothenuse c. 

The third angle, /3 = 90° — or. 

Logarithmic Solution. 

log a 0.95 424 cos a 9.90 309 

log b 1.07 918 log b 1.07 918 

tan a 9.87 506 log c 1.17 609 

a, 36° 52 7 .2 c, 15 metres. 

1 /?, 53° r.s 

* It is recommended that in commencing tnis subject the student first solve a 
few of the problems by his own process, and without the use of any tables but 
those he may construct for himself by measurement as described in § 18. 



40 



PLANE TRIGONOMETRY. 



1. 


Given a = 


2. 


" a = 


3. 


" a = 


4. 


" «== 


5. 


" a = 


6. 


" a = 



4, 



EXEECISES. 
5 = 



5; find remaining parts. 

5; 
84.107; " 
876.59: « 



43.148, 5 = 

2.7938, 5 := 
759.28, 5 = 51.85; 
8628, 5=27316; 

The first two exercises are made purposely simple, that they 
may be performed by measurement. 

38. Case II. Given the hyjpothenuse and one side. 
Solution. From the equation 

c sin a = a 
a 



we obtain 



sin a 



which may be used to find a when a and c are given. Then the 
remaining side is found by the equation 

b — c cos a. 

Example. Given a = 13, c = 20, to find the remaining parts. 
Solution by numbers and measurement. 

a 13 

20 



sin a = — = 



0.65. 



Take a distance equal to 0.65 of OX'm the dividers, and find 
out what, angle it will fit in the diagram (§ 15) to form a sine of 
an angle. This angle will be a. Measure off cos a perpendicu- 
larly to OB, take its ratio to OX, and multiply it by c = 5. This 
will give the side b. 

Logarithmic Solution. 

log. 13 1.11 394 cos a 9.88 078 

log. 20 1.30 103 log 20 1.30 103 

log b 1.18 181 

b, 15.199 



sin a 
a 



9.81 291 

40° 32'.5 

49° 27'.5 



OF RIGHT TRIANGLES. 



41 



Exercises. 
c= 9: 



find remaining parts. 



1. Given a = 7, ci 

2. " & = 9, c= 16; " " 

3. " a = 82.143, c = 120.412 ; " " " 

4. " b= 2.9235, c= 9.827; " " " 

5. A circle of radius r is drawn with its centre at a distance 
jp from a straight line. What length will it cut out from the line % 
What will be the result if r < p ? 

39. Case III. Given an angle and one side, as a and a. 
Solution, The equations (a) give 

z a 

o = =z a cot a = a cot p ; 

tan a: 



a 



c = 



sin « 



= # cosec a = a sec yff. 



Note. One angle being given, the remaining one may be found from the 

equation 

0=90° -a. 



Exercises. 



1. 

2. 
3. 
4. 
5. 



find a and b. 



C 



:*-P 



Given a — 72° 39', c = 19.5 ; 
" <* = 16° 25 / .6, c = 10.925 , 
« /? = 43° 28 / .4, J = 8.1273 ; find a and c. 
" /3= 8° 29'.2, 5 -- 0.9271 ; " " 
An engineer, desiring to find 
the distance from a point A on one ± 
bank of a river to a point P on the 
other bank, measured off a base line 
AC b yards in length, in a direction 
perpendicular to AP. He then 
measured the angle A CP, and found 
it to be a. What are the expressions for the distances AP and 
CP% 

What will be the distance if A C = 80 yards and or = 85° 22'.5 ? 
6. An engineer, desiring to determine the height of a vertical 



42 



PLANE TRIGONOMETRY. 



wall, measured off a distance PO = h feet on level ground, and 
then from a point E, a feet above 
the ground, measured the angle 
HEX = a between the line of 
sight EX and the horizontal 
line EH. Express the height 
PX of the wall algebraically in — ___ . 
terms of a, o, and ar, and com- 
pute the height when b = 400 
feet, a = 6 feet, and the angle a = 22° 17'. 

Method of solution. Find the height iZX and add HP = a. 

7. Desiring to find the height of an inaccessible rock M above 
a plane, its angle BPM M 

was measured from a 
point P and found to 
be a. The observer then 
advanced a metres to- 
wards the rock to Q, and ^ 

there again measured the P ^ Q a? B 

angle of elevation and found it to be y. Express the height BM 
of the rock. 

Method of solution. Let the vertical height PM= h and 

QB = x. Then we have the two right triangles PBM and 

QBM, which give 

h = (a -\- x) tan a ; 

h = x tan y. 

From these two equations we obtain the following values of h 
and x: 



sz. 




x = 



a tan a 



tan y — tan of 

-. a tan a tan y 

tb = -. 

tan y — tan a 

Find the height when a — 2000 yards, a = 30° 28', and 
y — 40° 53'. 

8. The altitude of a triangle is 7.2648, and the angles at the 
base are 72° 29.3' and 40° 30.5' respectively. Compute the base 



OF EIGHT TRIANGLES. 43 

and sides. Also, find the general expression for the length of the 
base in terms of altitude and angles at base. 

9. From the top of a tower 108 feet high the angles of de- 
pression of the top and bottom of another tower standing on the 
same horizontal plane are found to be 28° 56' and 53° 41/ respec- 
tively. Find the distance between the towers, the height of the 
second tower, and the distance between the summits of the two 
towers. 

10. In a circle of radius r, express the length of each side, and 
of the apothegm, of a regular inscribed polygon of n sides. Find 
first the special values for the triangle, square, pentagon, hexagon, 
and octagon. Ans. For the octagon : side = 2r sin 22^°; 

apothegm = r cos 22^-°. 

11. If the side of a regular octagon is 10 metres, what are the 
radii of the inscribed and circumscribed circles? 

12. At what altitude is the sun when a tower 20 metres high 
-casts a shadow 75 metres long upon a horizontal plane ? 

13. It was found that the length of the shadow of a monument 
upon a horizontal plane diminished 22 metres when the sun's alti- 
tude increased from 30° to 45°. What was the height of the 
monument ? 

14. If /3 is one of the acute angles of a right triangle, and c its 
hypothenuse, express the altitude in terms of c and J3. 

15. Two lighthouses, each 30 metres above the sea and 500 
metres apart, are seen by a ship in line with them to differ 1° in 
elevation above the horizon. "What is the distance from the 
nearer, supposing the ocean a plane ? 

16. The great pyramid of Gizeh is 762 feet square at its base, 
and each side makes an angle of 51° 51' with the horizon. Find — 

(a) Its height if continued to its apex. 

(b) The slope of its edges. 

Actually, instead of being continued to its apex, it terminates 
in a platform 32 feet square. Find — 

(c) The perpendicular height of this platform above the base. 
{d) The length of each edge from the corner of the platform to 

the corner of the base. 



CHAPTEK IV. 

RELATIONS BETWEEN FUNCTIONS OF SEVERAL ANGLES. 



The Addition and Subtraction Theorems. 

40, Problem. To express the sine and cosine of the sum of 
two angles in terms of the sines and cosines _. 

of the angles themselves. 

Solution. Let XOM = a and MOJV= ft 
be the two angles. XOJV = a -\- ft is then 
their sum. 

Let ON be the unit radius. From N drop 
NMiOM; NQlOX 
Prom M drop 

MPlOX; MPiNQ. ° 

Then sin (a + ft) = NQ = NR + MP; r 




cos (a + ft)=OQ = OP- MR. 
OQS and NMS being right angles, we have 

Angle RNM = comp. R8M= comp. OSQ = SOQ = a. 

NM= sin ft; 

NR = JVM cos a = sin. ft cos a ; 

OM = cos ft; 

M<P = OM sin a = cos ft sin or; 
whence, from (1), 

sin (a -f- ft) = sin a cos ft -\- cos a sin ft. 
We find in the same way 

OP = OM cos a = cos ft cos a ; 

i?_zlf/ == JVM sin ar = sin ft sin or ; 
whence, from (1), 

cos (a -f- ft) = cos a cos /? — sin a sin ft. 



(2) 



(3) 



THE ADDITION AND SUBTRACTION THEOREMS. 



45 




The formulae (2) and (3) constitute the addition theorem of 
trigonometry. 

To find the corresponding subtraction theorem it is only neces- 
sary to change the sign of ft. We have 

sin (— ft) = — sin ft; 
cos (— ft) = cos ft. 
Therefore, changing ft to — ft in (2) and (3), we find 

sin (a — ft) = sin a cos ft — cos a sin ft ; (4) 

cos (a — ft) = cos a cos ft -f- sin a sin /?. (5) 

It is, however, interesting to show how these equations may be 
obtained independently by a geometrical construction. 
Let POM be the angle a, and NOM ^ 

the angle ft. Then 

PON = a-ft. 
Take ON for the unit radius and 

drop 

NPlOP; NMiOM; 

MQiOP; NPiMQ. o 

Then sin (a - ft) = PN = QB = MQ - MB; 

cos (a-ft)=OP = OQ + PN. 

Because OMN is a right angle, 

NMR = comp. OMQ = Jf 6><0 = a ; 

ifiV = sin ft ; 

0Jf = cos ft ; 

Jfi? = J/i^ cos NMP = sin /? cos tf ; 

jlf# = OM sin Jf#() = cos ft sin a? ; 

OQ = OM cos if (9$ = cos ft cos a; 

PN = MN sin NMP = sin /? sin or. 

Making these substitutions, we have the results (4) and (5) for 

sin (a — ft) and cos (a — ft). 

41. $m0 and cosine of twice an angle. If we suppose 
ft = a, we have from (2) and (3) expressions for the sine and 
cosine of the double of an angle, namely : 

sin 2 a = sin a cos a -\- sin a cos a, (6) 

or sin 2 a = 2 sin a cos a ; 

cos 2 <* = cos 3 « — sin 2 a = (cos a -f- sin or) (cos a: — sin a). (7) 



46 PLANE TRIGONOMETRY. 

Also, by putting for cos 2 a its value, 1 — sin 2 a, and vice versa, we 

have 

cos 2a = 1 — 2 sin 2 a = 2 cos 2 or — 1. (7') 

EXEECISES. 

1. Because 90° = 60° + 30°, we have, by putting a = 60° and 
P — 30° in the equation (2), 

sin 90° = sin 60° cos 30° + cos 60° sin 30° = 1. 
It is required to test this equation by substituting the numeri- 
cal values of the sines and cosines of 30°, 60°, and 90° (§ 30), and 
to test in the same way the equations obtained by putting a = 60° 
and J3 = 30° in (3), (4) and (5). 

2. Because a = a — x -f- a?, we have, from (2), 

sin a = sin (a — x) cos x -f- cos (a — x) sin x. 
It is required to write the corresponding equations obtained by 
making the same substitution in (3), and the equations obtained in 
the same way from (4) and (5) by the identical equation 

a = a -\- x — x. 

3. Derive the addition theorem for the cosine from that for 
the sine by substituting y — 90° or 90° — y for /? in the equation 
(2), and applying the equations (18) of § 29 (Chap. II.). 

4. By means of the addition theorem prove the equations 
sin (a -\- /3 -j- y) = sin a cos /? cos y -f- cos a sin fi cos y 

-f- cos a cos fi sin y — sin a sin f3 sin y ; 

cos {a -\- fi -\- y) = cos a cos fi cos y — sin a sin /? cos y 

— sin a cos /? sin y — cos a? sin /? sin ^. 

Note. This is readily done by putting a -\- /? for a, and y for /J, in the 
equations (2) and (3), thus giving 

sin (a -\- (5 -f- y) = sin (a -f- /?) cos y + cos (a + fi) sin y ; 
cos (a -\- (3 -{- y) = cos (a -f /5) cos ^ — sin (a -f- ft) sin y ; 
and then developing by the addition theorem. 

5. Prove the following values of sin 3a and cos 3a : 

sin 3a = 3 sin a cos 2 « — sin 3 a ; 
cos 3 a = cos 3 a — 3 sin 2 <* cos or. 

6. Transform the expression 

# cos ( a + a?) + 5 cos (/? -}- #) + c cos (^ -f~ a?) 
into 



RELATIONS BETWEEN FUNCTIONS OF SEVERAL ANGLES. 47 

cos x (a cos a-\-b cos ft -\- c cos y) 
— sin x {a sin a -\- h sin ft -j- c sin y). 
7. If we have 

a cos oc -f- 1) cos /? -f- c cos y = 
and & sin ar -|~ Z> sin /? -|- g 1 sin y = 0, 

prove that we must also have 

& sin (a -\-x) -\-h sin (/? -\- x) + e sin (y -\- x) = 0, 
whatever be the value of x. 

42. TAtf Addition Theorem for tangents. Dividing equa- 
tion (2) by (3), we have 

sin (a 4- ft) , ^ sin a cos /? 4- cos a- sin /? 

; T ,i = tan {a + /?) = £-£- , r-C 

COS (a -\- ft) \ 1 / cos a cos ^ _ sm a sin ^ 

Dividing both numerator and denominator of the last member by 
cos a cos ft, the equation becomes 

, ^ tan a 4- tan ft 
tm ^ + /0 = 1 _J g J> (8) 

We obtain in the same way from (4) and (5), 

tan a — tan /? 
to («-fl = IT — ^ (9) 



Putting ft = a in (8) we have 

^ — 

1 — tan 2 a ' 



2 tan a 
tan2a = — — — . (10) 



Exercises. 

1. Assuming /? = 180°, prove by (8) of this section that 

tan (a + 180°) = tan a. 

2. Assuming a = 30°, substitute in (10) the value of tan 30° 
given in § 30, and thus obtain the value of tan 60°. 

43. Products of sines and cosines. Taking the sum and 
difference of equations (2) and (4) of § 40, and reversing the 
members of the equation, we find 

2 sin a cos ft = sin (a -\- ft) _|_ sin (a — ft) ;) 
2 cos a sin ft = sin (a -J- ft) — sin (a — ft). ) 



48 PLANE TRIGONOMETRY. 

In the same way we find from (3) and (5), 

2 cos a cos ft = cos (# + /?) + cos (a — /?) ; ) 
2 sin a sin ft = — cos (<* + /?) -f- cos (or — /?). ) 

Exercises. 

1. Prove that if a — /S = 90°, 

then cos {a -f- /?) = 2 cos a cos fi = sin 2«r = — sin 2/?. 

2. Prove that if a + = 180°, 

then sin (a — •. /3) = 2 sin a cos ft = sin 2/3 = — sin 2ar. 

44. $wm of sines and cosines. If in the four equations (11) 

and (12) we put 

ot-\- fi = x 

and a-/3 = y, 

we shall have 

and /? = i(a? — y). 

By substituting these values in (11) and (12) and reversing the 

members of the equation, we find 

sin x -\- sin y = 2 sin £(a? -f- y) cos -J(a? — y) ; " 
sin a? — sin y = 2 cos -J-(a? -f- y) sin £(a? — y) ; 
cos a? -}- cos y = 2 cos -J(a? -f- ^/) cos J(a? — y) ; 
— cos x -(- cos y = 2 sin -§-(# -f- ?/) sin |{a? — y). 

Dividing the first of this group of equations by the third, we get 

sin a? -f- sin y ... 

-r = tan i(» + 2/)- ( 14 ) 

cos x -\- cosy &\ \ yj \ i 

Dividing the second by the fourth, we get 

sin x — sin y ^ , , . , . K . 

- = cot *(a? + y). (15) 

cos 2/ — cos a? &\ \ v) \ j 

If in the last two equations of (13) we suppose y = 0, which 

makes cos y = 1, they become 

1 + cos a? = 2 cos 2 £a?; |_ . 

1 — cos x = 2 sin 2 ix;\ 

a pair of equations which frequently come into use. 

If we put a = Ja?, these equations become 

1 -f- cos 2a = 2 cos 2 a 

1 — cos 2a = 2 sin 2 a 

which may also be derived directly from (T ; ). 



(13) 



(16') 



(18) 



RELATIONS BETWEEN FUNCTIONS OF SEVERAL ANGLES. 49 

Dividing the second of (16) and (16') by the first, we find 

1 — cos x 

zr—. = tan 2 %x ; (17) 

1 -+- COS X * ' V ' 

1 — cos 2a 

—. g- = tan 2 a. (17') 

1 -f- cos 2a v J 

45. The problem of dimidiation. It is often required to 
express the sine or cosine of half an angle in terms of the sine or 
cosine of the entire angle. To effect this let us put in equations 
(6) and (7) 2a = y. They then become 

sin y = 2 sin \y cos \y ; 
cos y = cos 2 \y — sin 2 ^y, 
= 1-2 sin 2 \y, 
= 2 cos 2 \y — 1. 
Let our first problem be : 

Given, cos y, 
Required, sin \y. 
Solving the equation cos y = 1 — 2 sin 2 \y, we obtain 

i/l — cos V 
sin^ = |/ ^-. (19) 

Let our next problem be : 

Given, cos ^; 

Required, cos ^-^. 
Solving the equation cos y — 2 cos 2 \y — 1, we obtain 

i/l + COSV 

cosir = r 2 • (20) 

Let our third problem be : 

Given, sin y ; 
Required, sin \y. 
In the equation 

sin y = 2 sin -J/ cos Jy 

we put cos \y = Vl — sin 2 \y and square both members, obtaining 

sin 2 y = 4: sin 3 \y (1 — sin 2 \y). 

"Reducing, 

sin 2 y 
sin 4 \y — sin 3 \y + — j— = 0. 

Considering sin J^ as an unknown quantity, this equation can 



50 PLANE TRIGONOMETRY. 

be solved as a quadratic. Transposing the last term and adding 
J to each member, it becomes 

• 41 -si ,1 1 — sin8 Y cos3 Y . 
sm 4 \y - sin 2 \y + £ = ^-^- = — £— ; 

extracting the square root of both members, 

• o - - cos y 

sm J r - J = ± -^-; 

whence, by solving, 



sm ir = r — 2 _z -- 

Let our fourth problem be : 

Given, sin y ; 

Required, cos |-^. 
In the equation 

2 sin |-x cos \y = sin ^ 

we put sin \y = Vl — cos 2 %y, and then proceed as before. 
Solving with respect to cos \y, we shall find 



(21) 



V 1 =f= cos y , . 

cos£r = ^ £. (22) 

We have now to study the different values which these expres- 
sions for sin \y and cos \y may have in consequence of the double 
signs of the surds and the double signs under the radical sign. 

Equations (19) and (20) show that if the cosine of an angle y is 
given, the sine and cosine of half that 
angle may have either of two opposite 
values. The geometrical explanation 
of this is that to cos y may correspond 

either of two angles, y or 360° — y. ^*\>\'' \ ~"jSb* 

The halves of the general measure of 
y are \y and \y + 180° (§ 10). The 
halves of the general measure of 360° — y are 180° — \y and — \y 
(see diagram). 

Therefore the half angle may have either of four values. But 
because sin (180° -f- iy)= sin (— \y\ and sin (180° — \y) = sin \y, 
the sines and cosines of these four angles will have only two differ- 
ent values. This result agrees with equations (19) and (20). 




RELATIONS BETWEEN FUNCTIONS OF SEVERAL ANGLES. 51 



may 



But suppose the sine of y to be given. The angle y 
then have either of two supplementary 
values, y and 180° — y. The halves of 
the general measure of these angles will 
be \y, 90° - iy, 180° + \y , and 
270° -f- \y. The sines and cosines of these 
four angles are all different. Therefore ^° 
the algebraic expression for the sines 
ought to be susceptible of four values 
instead of two, as in the first case. Now this is true of equations 
(5) and (6), because they show that 

sinfr = + y ^o 




or 



-*A± 



cos y 



or 



+ 4 / i -cos Zor _ 1 /l-^s Z: 



and 



cosjr 



J\ — cos y 
= f 2 



or 



, i/i + cosr 

^ r 2 



or 



or 



i/ 1 — cosy 

r 2 



,/ l + cos y 
V ' 2 * 



Exercises. 

1. We have already found sin 30° = \ ; from this find the 
sine, cosine, tangent, and cotangent of 15° and of 75°. 

2. From the values of the six trigonometric functions for 45° 
(§30) find those for 22i°. 

3. From the values for 18° find those for 36°. 



46. Miscellaneous relations. The following equations are of 
occasional use in the applications of trigonometry, and can all be 
derived from the formulae of the last two chapters. Their deriva- 
tions are therefore presented as an interesting exercise. 
1. sin (45° - x) = cos (45° + x). 

COS X 



2. 


sin x = — — . 
cot X 






3. 


sin x = sin (60° + x) - 


- sin (60° - 


-x). 



52 PLANE TRIGONOMETRY. 

2 tan %x 



19. 



4. 

5. 


sin sd — i ■f. 2 -i • 
1 -f- tan 2 %x 

2 




~ tan -J^ -{-cot-Jo?' 




6. 


sin (30° + a?) — sin (30° - 


-X) 


sin x — ■ ■ .— 




7. 


sin x = 2 sin 2 (45° + Ja?) - 1 = 


1 - 


8. 


cos f = sin s cot £ . 
1 — tan 2 le 





2 sin 2 (45° - %x). 

= sin € cot £ . 

1 — tan 2 &e 
9. cos s 



10. cos # = 



1 + tan 2 -K 
cot \a — tan -§# 
cot \a -f- tan -J^* 
1 



11. 
12. 
13. 

14. 

1 K 


COS& — ^ . ;-■ ■ , ■ -, • 

1 -f- tan a tan J# 
cos a = 2 cos (45° -f £«) cos (45° - 

2 


-44 


C0 ° * ~" tan (45° + \a) + tan (45° - 
2 cot \e 

tan6?= cot 2 ^-r 

2 
tan — 


-M 


16. 
17. 


cot £0 — tan \& 

2 cot 2# = cot a — tan #. 

sin 20 
tan = — ■: ^. 





tan (45° + Ja) - tan (45° - 
tan a = 



18. ___ 



sin (^4 -|- -S) tan J[ + tan B cot i? + c °t A 
sin (A — B) ~ " tan ^4 — tan ./? ~~ cot i? — cot A' 
cos ( J. -f- B) _ cot i? — tan A _ cot J. — tan B 
cos ( A — B) ~ cot B -\- tan A ~ cot A -{- tan i?' 

21. cos a cos J = cos 2 \{a -\- ~b) — sin 2 \{a — h). 

22. sin a sin J — cos 2 %(a — h) — cos 2 J(a -f- b). 

23. tan # -|- cot a = 2 cosec 2#. 

24. sin 2 <z cos 2 /? — cos 2 a sin 2 j3 = sin (or -j- /?) sin (<*—/?). 

25. cos 2 a: cos 2 /? — sin 2 a sin 2 /? = cos (a -f- /?) cos (a— ft). 



CHAPTER Y. 

TRIGONOMETRIC PROBLEMS. 



47. Problem I. Hawing given two equations of the form 

r sin cp = a, 

r cos cp = b, 

where a and b are given numbers, it is required to compute r 

and cp. 

Solution. Dividing the first equation by the second, we find 

sin cp a 

= tan cp = T . 

cos cp o 

From this value of tan cp we find cp itself, then the sine or 
cosine of (p, and then r from either of the equations 

a b 

sin cp cos cp 
Example. If r sin cp = 332.76, and r cos cp = 290.08, it is 
required to find r and cp. 
log sin <p, 9.877 25 (5) 
" r sin <p, 2.522 13 (1) 
" r cos cp, 2.462 52 (2) 
" r = log r sin cp — log sin cp, 2.644 88 
" r = log r cos cp — log cos <p, 2.644 88 

r, 441.45 

The numbers in brackets show the order in which the num- 
bers of the computation are written. In writing log r sin cp and 
r cos cp spaces are left for inserting log sin cp and log cos cp after cp 
is found, so that either of the latter may be subtracted to obtain 
r. It is generally best to obtain r from both r sin cp and r cos cp, 
because then if any mistake is made in cp it will be shown by a 
difference in the results. 



log COS 


% 


9.817 64 


(6) 


" tan 


9, 


0.059 61 


(3) 




9, 


48° 55'.2 


(4) 


(1) 








(8) 








(9) 









54 PLANE TRIGONOMETRY. 

On the other hand, a practical computer will not write down 
either sin cp or cos (p, but will subtract in his head and write down 
log r only. 

Exercises. 

1. Given r sin cp = 1.297 43, r cos cp = 6.002 4 ; find r and cp. 

2. " " =0.08219, " =0.128 8; " " 

3. « " =194 683, " =8460.7; " " 

48. Distinction of quadrants. In the preceding examples 
we have supposed r sin cp and ^ cos cp to be positive, and have 
taken cp in the first quadrant. But either or both of these quan- 
tities may be negative. Whatever their signs, there are always 
two values of cp, differing by 180°, corresponding to any given 
value of tan cp (§ 31). Hence the problem admits of two solu- 
tions in all cases. In the one r will be positive, in the other 
negative. 

But in practice only that solution is sought which gives a posi- 
tive value of r. This being the case, sin cp and cos cp must have 
the same algebraic signs as the given quantities r sin cp and r cos cp 
respectively. ISTow consider each case in order : 

I. ?* sin cp and r cos cp both positive. The angle cp must then 
be taken in the first quadrant, because only in this quadrant are 
sin cp and cos cp both positive. 

II. r sin cp positive and r cos cp negative. Sin cp is positive 
only in quadrants (1) and (2) (§ 21), and cos cp is negative only in 
quadrants (2) and (3). Hence the requirement of signs can be 
fulfilled only in the second quadrant, and 

90°<<p<180°. 

III. r sin cp and r cos cp both negative. The only quadrant in 

which sine and cosine are both negative is the third. Hence in 

this case 

180° < <p<270°. 

TV. rsin cp negative and r cos cp positive. The only quadrant 

in which sin cp is negative and cos cp positive is the fourth. Hence 

in this case 

270°<^<360°. 



TRIGONOMETRIC PROBLEMS. 55 

Exercises. 

1. Given r sin cp = — 237.09, rcos cp = -\- 192.91 ; find r and cp. 

2. " " + 2713.8, " - 9269.2 ; " 

3. " " -1.9634, " -0.09654; " " 

4. " " -3.6925, " +396.72; " " 

5. " " -1S.005, " -2.6943; " " 

6. Given 7-sin (9? + 47° 50') = 7.2693, 

r cos (<? + 47° 50') = - 12.2916, 
to find r and cp. 

Note. In this last exercise compute the value of (<p -|- 47° 50') as if it 
were one quantity, and subtract the angle 47° 50' from the result. 

7. Given r sin (d + x) = — 249.88, 

r cos (0 + x) = — 92.62, 
^sin(<9-^) = 702.02, 
ucos(6 — x) = 516.93; 
find the values of r, u, 0, and x. 

8. Given r (sin 6 + cos 0) = 298.07, 

r (sin 6 — cos 0) = 96.04 ; 
find r and 6. 

49. Problem II. Having given two equations of the form 
x cos a -j- y sin a = ^?, 

a? sin a: — y cos a = q 7 

it is required to find the values of x and y. 

The elimination is conducted by the method of addition and 

subtraction, as follows : 

Multiply the first equation by cos a and the second by sin a. 

We thus have 

x cos 2 a -f- y sin a cos a = p cos or ; 

x sin 2 <z — 3/ sin a cos a = q sin ar. 

Now adding these equations together and remembering that 

sin 2 a -f- cos 2 a = 1, 
we have 

a? = p cos a -f- ^ sin a, 

which is the required value of x. 



56 PLANE TRIGONOMETRY. 

Next multiply the first equation by sin a and the second by 

— cos a. We have 

x sin a cos a -\- y sin 2 a = p sin a ; 

— x sin a cos <* -f- y cos 2 a = — q cos or. 
By addition, 

y = ^? sin a: — ^ cos or. 

It will be noticed that in these equations x and y are given in 

terms of p and ^ by equations of the same form as the original 

ones. 

Exercise. 

Find the value of p and q from the equations 
p sin a -j- q cos a = a ; 
^? cos or — ^ sin a = 5. 

50. Problem III. From the equations 

r sin (/? -|- e) = a, 

r sin (/? + 6) = J, 

to find the values of r and /?, — the other four quantities, a, b, s > 

and 6, being supposed known. 

Solution. Developing the sines of fi -f- a and fi -j- 6 (§ 40), 

we have 

r sin /? cos £ -f- r cos /? sin £ = a ; ) 

r sin /? cos 6 -f~ r cos /? sin 6 = b. J ^ ' 

Regarding r sin /? and r cos /? as the two unknown quantities, we 
see that their coefficients are cos £, sin £, cos 6 and sin 6. Multi- 
plying the first equation by cos 6 and the second by cos e, we have 

r sin /? cos £ cos 6 -f- r cos /3 sin £ cos 6 = a cos 6 ; 

r sin /? cos £ cos & -\-r cos /3 cos £ sin = b cos £. 
Subtracting, 

r cos /? (sin £ cos 6 — cos £ sin 6) = a cos 6 — b cos £. 
Noting that the coefficient of r cos /3 is sin (£ — 6\ we find, by 
division, 

a cos — b cos £ 

r cos p = : — t ^ — • 

^ sm (e — 6) 

To find the value of r sin /3 we multiply the first equation (a) 

by sin # and the second by sin £, and subtract. We thus find 

r sin /3 (sin £ cos — cos £ sin 6) = 5 sin £ — a sin 6. 



TRIGONOMETRIC PROBLEMS. 57 

b sin s — a sin 6 
Hence r sin /* = sin (£ _ ^ • 

Supposing the numerical values of a, b, e, and 6 to be given, 
these equations give the values of r sin /? and r cos /?, from which 
r and /? can be computed by Problem I. 

Exercises. 
Find the values of r sin cp and r cos cp from the following 
equations by the preceding method : 

1. r cos (cp -\-s) = a, r cos (cp— 6) = J. 

J COS £ — (2 COS 

Ans - *" sm< P = sin(* + 0) ; 

J sin £ 4- a sin 

roo8 »'= sin( S + 0) • 

2. z 8 cos (cp -\- s) = a, r sin (^» + 0) = &• 

J cos e — asm 
Ans. nmffl = > ^ — ; 

^ COS (fi — C7) 

5 sin £ 4- a cos 
r cos cp = t ^r — . 

^ COS (£ — 0) 

3. r sin (<p + e) = a cos ^, r cos (cp — e) = a sin ^. 

a cos (y + £) 
Ans. rsmcp=- ~ '-• 

cos 2f ' 

6 sin (v — £) 
y cos cp = ^— \ 

cos2e 

4. Find the values of r and cp from the equations 

7>cos(<p4-0) = 3.790 8, 
r cos (cp — 6)= 2.060 7, 
when = 31° 27'.4. 

51. Problem TV, To reduce an expression of the forrr* 

a sin 6 -|- b cos 

&? # monomial, a and b being given quantities. 

Solution. Determine the values of two auxiliary quantities k 

and s from the equations 

Jc cos £ = #, 

&sin £=5, 
as in Problem I. 



58 PLANE TRIGONOMETRY. 

The given expression will then become, by substitution, 

k cos £ sin 6 -\-Jc sin e cos 6 = h sin (d -\- e). 

We might equally have supposed Tc sin € = a and h cos e = b> 
when the given expression would have become h cos (6 — e). 
Example. Reduce the expression 

1239.3 sin x — 724.6 cos x 
to a monomial. 

Jc sin s = — 724.6 
Jc cos s = 1239.3 



log = 


- 2.860 10 


log = 


3.093 18 


log tan s , 


- 9.766 92 


*, 


- 30° 18'.8 


log COS £, 


9.936 15 


log*, 


3.157 03 


Jc, 


1435.6 



We therefore have 

1239.3 sin x - 724.6 cos x = 1435.6 sin (x — 30° 18'.8). 

EXEECISES. 

Reduce to monomials the expressions : 

1. 27.615 cos m — 23.208 sin p. 

2. 3.600 3 sin (6 - 7° 52'.6) + 5.907 sin (0 + 53° 57'.6). 

3. Reduce to a monomial the expression 

92.65 sin 6 cos a — 196.23 cos 6 sin a, 
when a = 162° 48'.7. 

4. Reduce to a monomial the expression 

cos a cos -f- sin a sin 6 cos H, 
when a = 62° 39'.5 and H= 22° 36'.8. 

52. Problem Y. 7^ reduce an expression of the form 

a sin x cos -f- 5 cos a? sin (1) 

to a form which shall not contain the product of any two trigono- 
metric functions. 

Solution. We have, from § 43, 

sin x cos = \ sin {x -f- 6) -f- -J- sin (a? — 0) ; 
cos a? sin — J sin (as -f- 6) — \ sin (a? — 6). 



TRIGONOMETRIC PROBLEMS. 59 

Making these substitutions in the expression (1), it becomes 
i(a + b) sin (x+6) + #a - b) sin (x - 6) 

Exercises. 

Clear from products of sines and cosines : 

1. m cos a sin ft — n sin a cos ft. 

2. a cos 6 cos /* + ° sin sin pi. 

3. # cos 6 cos m — b shi 6 sin /*. 

53. Peoblem VI. From the equations 
r cos ft cos A — a, ^ 

r cos /? sin A = 5, I (1) 

/* sin /? = c, J 

to find the values of r, ft, and A, the values of a, b, and c being 
given. 

Method of solution. Dividing the second equation by the 
first we obtain 

tanA = -; (2) 

from this equation we find A, and then sin A or cos A from the 

tables. Then 

a b 

rco& ft = r = — — - (3) 

cos A sm A w 

can be computed. The value of r sin ft being given by the third 
equation (1), the values of r and ft are found by Problem I. 
Example. Find r, ft, and A from the equations 
r cos ft cos A = — 53.953 ; 
r cos /?sinA = -f 197.207; 
r sin ft = — 39.062. 

Work log cos ft, 9.992 21 

log r cos /?, 2.310 60 

log r sin ft, - 1.591 75 

log tan ft, - 9.281 15 

log r, 2.318 39 

r, 208.16 

ft, - 10° 49' 



log r cos /? cos A, - 
log r cos ft sin A, 
sin A, 


- 1.732 01 

2.294 93 
9.984 33 


log tan A, • 
log r cos /?, 


- 0.562 92 

105° 18'.0 

2.310 60 



60 PLANE TRIGONOMETRY. 

Exercises. 
Find the values of r, /?, and X from : 

1. r cos /? cos X = 1.271 83 ; 
r cos /? sin X = — 0.981 52 ; 
r sin /3 = 0.890 02. 

2. ^ sin /? sin A = 19.765 3 ; 
r sin /? cos X = — 7.192 8 ; 
rcos/3 = 12.124 2. 

Miscellaneous Exercises. 

1. Compute go and r from the equations 

1.268 22 sin go = 0.948 30 + r sin (25° 27'.2) ;, 

1.268 22 cos go = 0.281 16 + r cos (25° 27'.2). 

First eliminate r. 

Ans.. a? = 60° 53'. 8; 

r = 0.371 7. 

2. Find go and a? from the equations 

3 sin go -f- cos a? = 2a?. Ans. go = 71° 34'; 

sin a> -f" 2 cos 63 = x. x = V%. 

After finding sin go and cos go in terms of # employ the equation 

sin 3 go -f- cos 2 go = 1. 
Find a? from the following quadratic equations, and express 
the results without surds : 

2a? 

3. a? 2 + 1 = - — . Ans. a? = cot i a or tan ^ a. 

1 sin a ■* ■ * 

a a . 1 o a 1 ± sin a 

4. a? +1 = 2a? sec a. Ans. a? = 

cos a. 

5. 1 — a? a = 2 a? cot a. Ans. a? = tan £« or — cot \a. 

a a -i o j. a sin a ± 1 

6. ar — 1 = 2 a? tan or. Ans. a? = 

cos a. 

Find (9 from the equations : 

7. 27.615 cos — 23.208 sin 6 = 19.094. 

8. 3.6003 sin (0 - 8°) + 5.907 sin {6 + 54°) = 2.6253. 
Keduce the first members by Prob. IV. 

9. a sin 6 -(- J cos -f- <? = 0. 

From this last equation find sin 6, cos 0, and tan 6 by separate 
quadratic equations. 



CHAPTER VI. 
SOLUTION OF TRIANGLES IN GENERAL. 



54. A plane triangle has six parts, three sides and three angles. 
Of these parts the three angles are not independent of each other, 
because when two angles are given the third may be found from 
the condition that the sum of the three angles is 180°. Hence, if 
two angles are given, the case is the same as if all three were given. 

When any three independent parts are given the remaining 
three may be found, but in order to be independent one of the 
three given parts must be a side. 

55. The fact that the sum of the three angles of a plane 
triangle is 180° enables us to express a trigonometric function of 
any one angle as a similar function of the sum of the other two 
angles. It has been shown that 

sin x = sin (180° — x) ; 
cos x = — cos (180° — x) ; 
tan x = — tan (180° — x) ; 
cot x = — cot (180° — x) ; 
sec x = — sec (180° — x) ; 
cosec x = — cosec (180°— x). 
If a, ft, and y are the three angles of a triangle, we have 

« = i80°-(/? + r)n 

fi = 180°-(y + *);\ (a) 

y = 180° - (a + /J). J 



Therefore 



sin a = sin (fi -f- y) ; ' 
sin /? = sin (y + a) ; 
sin y = sin (a -f- ft) ; 
cos a = — cos {fi-\-y)\ 
etc. etc. 



(i) 



62 



PLANE TRIGONOMETRY. 



By dividing the equations (a) by 2 we find 
i/?= 90° -£(/ + «); 



Therefore 



sinja = cos%(/3-\-y) 
cos^a = sin J(/? + x) 
tan J <* = cot ^ (/? -|- 7) 
cot J « = tan J (/? + Y) 
etc. etc. 



(*> 



(2> 



From what has been said the given parts may be 

One side and the angles ; 

Two sides and one angle ; 

The three sides. 
Also, when the sides and one angle are given, this angle may be 
either that between the given sides or opposite one of them. 
Hence there are four cases in all to be considered. 

56. Case I. Given the angles and one side. 

Theorem I. The sides of a plane triangle are proportional 
to the sines of their opposite angles. 
Proof. Put 

a, b, c, the three sides ; 
a, /?, y, their opposite angles. 
From one angle, as y, drop a perpendicu- 
lar yD upon the opposite side, c. Then 
yD = 5sin a ; yD = a sin /3. (§ 35) 

Therefore 

a sin fi = b sin a. 

Dividing by sin a sin /?, 

a b 

sin a ~ sin /?* 
By dropping a perpendicular from a upon a we should find, in 
the same way, 

b c 

sin fi ~ sin y ' 




TRIANGLES IN GENERAL. 63 

a b e 

Therefore - — = - — a = - — , (3) 

sin a sin /3 sin y* v J 

or a : b : c = sin a : sin /3 : sin ;/. Q.E.D. (4) 

ZW\ The common value of the three quotients -: , -: — -, 

J ^ sin a' sin ft 

and - — is called the modulus of the triangle, 
sm y 

Theorem II. The modulus of the triangle is equal to the 
diameter of the circumscribed circle. 

This theorem may be demonstrated by the student from the 
property that an inscribed angle is measured by one half the arc 
on which it stands. 

Theorem I. enables us, when the angles and one side of any 
plane triangle are given, to find the remaining two sides. If the 
side given is c, we have 



csm a 
a = 



b = 



csin/3 



(5) 



smy 

or, putting M for the modulus, we have 

a = M sin a ; 1 

b = Jlfsin/?; \ (6) 

c = M sin y. J 
If we put p, p', and p" for the lengths of the perpendiculars 
from a, /?, and y respectively upon the opposite sides, we find, 
from the preceding figure, 

p = b sin y = c sin /? ; 1 
p' = c sin a = a sin y ; \ (7) 

p" = b sin a = a sin /?. J 
By these equations we may find the lengths of the perpen- 
diculars. 

Exercises. 

1. Given a = 78° 23'.2, p = 52° 16'.3, a = 796.25 ; find b and c. 

2. " «= 5° 26'.2,/? = 72° 36'.8, 6=19.263; " b and c. 

3. « a = 50° 58'.7, /? = 32° 50'.8, c = 169.37 ; " a and 5. 



64 



PLANE TRIGONOMETRY. 



B 




V~~C 



4. In order to find the distance of a point G across a river 
from the points A and B, a sur- 
veyor measured a base line AB 
and found it to be 829.72 metres. 
Placing a theodolite at A, he found 

Angle B A G = 82° 37'. 6. 
Carrying his theodolite to i?, he 
found 

Angle ABG = 70° 3'.3. 
Required the distances .A (7 and BG across the river, and the 
length of the perpendicular from G on AB. 

5. In a triangle Ai?£7 the angle B exceeds the angle A by 
10°, the angle G exceeds the angle B by 20°, and the side A G is 
2.72 904 metres. Find the angles and sides of the triangle and 
the length of the perpendiculars from the angles upon the oppo- 
site sides. ♦ 

6. The base of an isosceles triangle is 132.643 metres, and the 
angle at the vertex is 32° 53'.7. Find the sides and the altitude. 

7. One diagonal of a parallelogram measures 23 metres, and it 
makes angles of 32° 17' and 63° 24' with the sides. Find the 
lengths of the sides and the angles of the parallelogram. 

8. From a point at a distance a from the centre of a circle of 
radius r tangents are drawn to the circle. Express the lengths of 
the tangents and the distance between the points of tangency, and 
compute the result when r = 7, a = 12. 

57. Case II. Given two sides and the angle op- 
posite one of them. 

Let the given parts be a, b, a. We then compute the parts /?, y, 
and c by the formulae (a) and (3) already found. 

■ a h • 

sm/3 = -sin or; 



c = 



180° -(a + fi); 

bsiny &sin;r 



(8) 



sin/? ' ' sma 

This case may have two solutions, as is shown in geometry. 
The two solutions are found in the above equations, because to a 



TRIANGLES IN GENERAL. 



W 



given value of sin/? corresponds either of two angles fi (§21), 
which will be supplements of each other. 

But if one solution gives «-f /? > 180° it is not admissible, 
and only the lesser value is used to give the triangle. This will 
be the case when a > b. 

It may also happen that sin /3 = - sin a comes out greater than 

unity. There is then no possible triangle which fulfils the con- 
ditions. 

Example. Given a = 152.08, b = 236. 74, a = 32° 29'.6 ; find 
the remaining parts. 

log b, 2.374 27 

log sin or, 9.730 14 

co log a, 7.817 93 



log sin fi, 9.922 34 

/?, 56° 44'.9 
a, 32° 29'.6 

a + fr 89°14 / .5 
180° OO'.O 



or 



r , 90° 45'.5 



log sin y, 9.999 96 

log -^-,,2.45193 

& sin fi 



log o, 2.4518 
c, 283.07 



or 
or 



or 
or 



123° 15'.1 
32° 29'.6 

155° 44'.7 
180° OO'.O 

24° 15'.3 

9.613 62 

2.451 93 

2.065 55 
116.29 



find the remaining parts. 



Exercises. 

1. Given a = 24, b = 33, a = 31° 28' 

2. " a = 34, 5 = 35.79,/? = 17° 59' 

3. " a = 29, J = 34, a = 30° 20' 

4. " J =19, c = 18, r=15°49' 

5. " a = 24, c = 13, ar = 115°0 / 
6. From a point P at a distance a from the centre of a circle 

of radius r a line is drawn, making the angle /? with the line from 
P to the centre of the circle. At what distances from P will the 



ffi PLANE TRIGONOMETRY. 

line intersect the circle ? Compute the distances when r = 72, 
a = 98, and /? = 28° 56' '. 

7. Show that if, in the present case, we take the side which is 
not given as the base, we can find the altitude of the triangle im- 
mediately, and afterwards may find the three required parts of the 
triangle from the altitude. 

Case III. Given the three sides. 

Theorem III. In a triangle the square of any side is equal 
to the sum of the squares of the other two sides minus twice the 
product of these two sides into the cosine of the angle included 
by them. 

In symbolic language this theorem is expressed in any of the 

forms _ 9 , _ 

a 2 = b 2 -f- g — 2bc cos a, 1 

or b 2 = a 2 + c 2 -2ac cos /?, [ (9) 

or c 2 = a 2 -\- b 2 — 2ab cos y. J 

Proof. It is shown in geometry that in any triangle the side 
opposite an acute angle is greater than the sum of the squares 
of the other two sides by twice the product of one of these sides 
into the projection of the other side upon it. 

If a be the acute angle, we have, by this theorem, 
a 2 = b 2 -f- c 2 — 2b X (projection of c on b). 
By §35, II., and the definition of projections. 

Projection of c on b = c cos a ; 
substituting this value of the projection, 

a 2 = b 2 + c 2 — 2bc cos a. Q.E.D. 

The other equations may be proved in exactly the same way. 

If the containing angle is obtuse, the square of the opposite 
side will be greater than the sum of the squares of the containing 
sides. But this case is included in the trigonometric formula, 
because then cos a is negative and — be cos a is positive. Hence 
the formula is applicable to all cases when regard is had to the 
algebraic sign. 

From the first of equations (9) we obtain 

C0 * a= Wc ' (1 °> 



TBIANQLE8 AY GENERAL. 07 

which with the two companion formulae enable us to find the 
angles when the three sides are given. 

59. If the angle is small, it cannot be accurately determined 
by means of its cosine ; we therefore transform the expression as 
follows : 

Subtracting each member of the equation from unity, we have 

V + tf-a 1 ' 2bc-b*-c* + a* 
l_ cos „= i _ -—r = _ . 

But 1 - cos a = 2 sin 2 \a (§ 44), and 2bc -b*-c 2 =-(b — c)\ 
Therefore 

J sin 2 a- 2ho - 2bc 

Let us now put s for half the sum of the three sides, so that 

s = i(a + b + c). 
Then a + b — c = 2s — 2c, 

a-b + c = 2s-2b; 
and the preceding equation reduces to 

sin H « = ( *-^- C) . (11) 

The expressions for the other two angles are obtained by the 
same process, the letters a, b, c and a, /?, y being permuted in the 
orders I, c, a ; fi, y, a and c, a,b ; ^, or, j3. We thus find 

(5 — c) (s — a) 



2 ca 

. (s — a)(s — b) 

*™*ir = - s ■ 



(12) 



These equations answer our purpose, but in determining an 

angle the tangent is the function to be preferred, because an angle 

can be determined more accurately from its tangent than from its 

sine or cosine. To obtain expressions for the tangent add unity 

to both sides of the equation (10). We then have 

„ , b 2 + c 2 -a 2 b 2 + 2bc + c 2 -a 2 
l + CO s« = 1 + _ s __ = _ . 

Since 1 -|- cos a = 2cos 2 \a (§ 44), this equation reduces to 
2cos ia = = . 



68 



PLANE TRIGONOMETRY. 



Whence 



cos 2 Jo: = 



s(s — a) 



(13) 



Dividing (11) by this equation, and writing the corresponding 
equations for the other angles, we find 

(s-b)(s-e) 

s(s — a) 



tan 2 \ a = 



tan 2 £/? 



tan 2 £/ = 



(s — g){s — a) 

(s — a)(s — b) 
s(s —c) 



(14) 



The computation will be simplified a little by computing 

E ^ a A* -a)(s- b) (s - c) 
s 
We shall then have 

tan J a = 



H 



tan J /3 = 



tanj^ = 



(•-a)' 

(s-*) ; 



(15) 



By means of these equations we may compute two of the 
angles, and find the third by subtracting their sum from 180°. 
But in practice it is better to compute the three angles indepen- 
dently, and check the accuracy of the work by taking their sum. 

If this sum comes out materially different from 180°, there is 
some mistake in the work ; if not, it may be presumed correct. 

Example. Given a = 273.960, I = 198.632, c = 236.914 ; 
find the angles. 



a = 273.960 

I = 198.632 

c = 236.914 

2s = 709.506 

,9 = 354.753 

s—a = 80.793 

s - 1 — 156.121 

s- c = 117.839 



log (s -a), 1.907 37 

log -b), 2.193 46 

log (s -c), 2.071 29 

sum of logs, 6.172 12 

log s, 2.549 92 

log R\ 3.622 20 

log H, 1.811 10 



log tan Jar, 9.903 73 
log tan J/?, 9.61764 
log tan ^,9.739 81 



TRIANGLES IN GENERAL. 69 



Ja, 38° 42'.1 ; 


a = 


77° 24'.2 


£/?, 22° 31'.2 ; 


/» = 


45° 2'.4 


Jy, 28° 46'.8 ; 


r = 


57° 33'.6 



Sum = 180° 0'.2 (Check.) 
The discrepancy of 0'.2 is the result of the unavoidable errors 
from the omission of the decimals of the logarithms beyond the 
fifth. 

Another check on the accuracy of the work is obtained by 
computing the modulus of the triangle from its three separate 
expressions (§ 56, 3), and noting whether they agree, thus : 

log a, 2.437 69 log b, 2.298 05 log e, 2.374 59 

sin a, 9.989 42 sin /?, 9.849 79 sin y, 9.926 32 

log modulus, 2.448 27 2.448 26 2.448 27 

The three results agree within the unavoidable limits of error. 

Exercises. 

1. Given a = 3, 5 = 4, <? = 5 ; find the angles. 

2. " a = 37 593, b = 29 867, c = 40 005 ; " " 

3. « a = 2.796 1, b = 23.928, c = 25.046 ; " " 

4. The base of a parallelogram is 13, each side is 6, and it& 
lesser diagonal is 12. Find its angles. 

5. If the sides of a parallelogram are a and b, and one diagonal 
is^>, express its angles. 

6. The parallel sides of a trapezoid are 12 and 17, and the non- 
parallel sides 6 and 7. Find its four angles. 

Suggestion. Divide the trapezoid into a triangle and a parallelogram. 

7. In a triangle are given the two sides, 
p and <?, and the medial line r from the 
vertex to the middle point of the base. 
Prove Base = V2p* + 2q* — 4r 8 . 

8. Given the three medial lines r, r', r n of a triangle ; find 
the sides of the triangle from the preceding result. 

Ans. | i/2r 8 + 2r" — r' n ; f V2r 2 + 2r" 2 — r"; f V2r' 2 + 2r"*— r\ 

9. Of the bisectors of the angles at the base of a triangle, the 
one cuts the opposite side in the ratio m : n, and the other in the 




70 



PLANE TRIGONOMETRY. 



Ans. cos a = 




ratio p : q. By means of the equation (10) express the cosine of 

the angle at the vertex of the triangle. 

jfn* -f- q*m 2 —p^m* 
2pqmn 

Note. In the solution of this question apply 
the theorem of geometry which defines the ratio in 
which the bisector of an angle cuts the opposite side. 

60. Case IV. Given two sides and the included 
angle. 

This case may be readily solved by Theorem III., because if 
the given sides are b and c, and the given included angle is tf, we 
have for the third side 



a = Vlf + £ 2 — %bc cos a. 
Then, having the three sides, the remaining angles may be found 
as in the last section. But there is a more convenient method 
founded on the following theorem : 

Theorem IV. As the sum of any two sides 
is to their difference, 
so is the tangent of half the sum of the angles opposite these sides 
to the tangent of half their difference. 
Proof From the equation 

b : c : : sin ft : sin y, (Th. I.) 

we have, by composition and division, 

b + c : b — c : : sin ft -f- sin y : sin ft — sin y. 

But sin /? + sin y = 2 sin i(ft + y) cos i(ft - y); (§44) 

sin ft — sin y = 2 cos i(ft -f- y) sin \{fi — y). 

Substituting these values, and expressing the proportion as a 

fraction, 

b -f- e _ sin j{ft -f- y) cos j(ft — y) 

b — c ~ cos i(ft -f- y) sin i(ft — y) 

= tan Hfi+y)QQkHfi-f) (§33) 

_ taniQg + y) 
tan i(ft — y)' 



Therefore 

b-\-c :b — c 



timi(J3 + y):tmi(ft-y). Q.E.D. (16) 



TRIANGLES IN GENERAL. 71 

The solution is now obtained as follows : "We have 

K/*+r)=9o°-i« 5 

tan #J3 +y)= cot Ja. (§ 55) 

Because the angle a is given, the only unknown term of the 
proportion is tan £(/? — y). This is given by the equation 

tan i(fi -y)= |^ tan ftf + r ), (17) 

which is derived from the proportion (15). 

By this equation we obtain £(/? — y\ which being added to 
and subtracted from £(/? + Y) gives /3 and y. The remain- 
ing side of the triangle may then be found by Case I. But 
when this side as well as the angles are required, a more elegant 
method may be followed, which will be explained in the next 
section. 

Example. Given b = 4.567, c = 3.456, a = 56° 7'.8 ; find the 
remaining parts. 

180° 
a, 56° 7'.8 





b, 4.567 




c, 3.456 




b - c, 1.111 




b+c, 8.023 




log (b - c), 0.045 71 




colog(5 + c), 9.095 66 


log 


tan i(/3 + y), 0.273 14 



log tan i(/3 - y), 9.414 51 



+ y, 123° 52'.2 
W + y), 61°56'.l 
4(/J - r ), 14° 33'.6 

A 76° 29'.7 
^, 47° 22'.5 



log 5, 0.659 63 log c, 0.538 57 

log sin fi, 9.987 82 log sin y, 9.866 76 

log Mod., 0.671 81 log Mod., 0.671 81 

a = 3.899 8 log sin a, 9.919 23 

log a, 0.591 04 

Exercises. 

1. Given a = 12.34, b = 43.21, y = 34° 12'; find rem. angles. 

2. " 5 = VZ, c = V3, a = 35° 53'; " " 

3. " a = 35.79, c = 1.246 8, /? = 97° 53' ; « " 

4. " a = 189, b = 114.75, ^ = 107° 48' ; « " 



(•> 

am 

(§55) 
(b) 



72 PLANE TRIGONOMETRY. 

61. If in the present case all three remaining parts are 
wanted, formulae may be derived as follows : 
From the equations (3) we derive 

b _ sin J3 c _ sin y 

a ~ sin a ' a sin a * 
Adding these equations we have 

b + e _ sin j3 -f sin y _ 2 sin -§-(/? + y) cos j(/? — y) 
a ~~ ' sin a sin a 

2 cos ja cos j(/? — /) 
2 sin |-« cos -Ja 

cos j(/? — r) 

sin ^« 
Subtracting the equations (#) we have 

& — c_ sin yff — sin y _ 2 cos J(J3 -f ^) sin -|(/? — ^) 
#"" "~ sin a 2 sin ^a cos \ol 

COS f« v J 

From the equations (c) and (J) we obtain 

a sin £(/? — ^) = (b — 6) cos £a, ) 
# cos £(/? — r) = 0& + c) sin i« ; J 
which equations are readily solved by Prob. I. Chap. V. 

By taking the quotient of these equations we may readily 
deduce the relation (16). 

Exercises. 

1. Given £=2956.2, c=9090.8, a=9S° 29'.6; find /?, y, and a. 

2. A surveyor lays off two lines from the same point : the one 
due north, 279.25 metres, the other east 15° north, 109.262 metres. 
How far apart are the ends of the lines, and what is the direction 
of the line joining them ? 

3. The sides of a parallelogram are 26 and 15, and one angle is 
126° 52'.2. Find the lengths of the two diagonals and the angles 
which they make with the sides. 

4. Given the two diagonals d and df of a parallelogram and 
the angle e which they form ; express the sides and angles of the 
parallelogram algebraically and compute them for the special case 
d = 5, d' = 6, s = 49° 18'. 




TRIANGLES IN GENERAL. 73- 

Areas of Triangles. 

62. Theorem V. The area of a triangle is equal to half the 
product of any two sides hy the sine of tJieir included angle. 

Proof. It is shown in geometry 
that the area is half the base into the 
altitude. Now in the figure, 

Altitude h = o sin y. 
Therefore, a being the base, 7 ^ a x ^ 

Area = \ah = \ah sin y. Q.E.D. 

Cor. 1. If two triangles have two sides of the one respectively 
equal to two sides of the other, and the angles which these sides 
form supplementary, the triangles will he equal in area. 

For the sines of the supplementary angles are equal. 

Cor. 2. Since we may take any one side as a base, if we call 
h, h' , h" the altitudes above the respective sides a, h, and c, we 
shall have a7l = W = ch ^ 

and db sin y = he sin a = ca sin /?. 

For these expressions are each double the area of the triangle. 

Exeecises. 

1. Given a = 75, I = 29, /? = 16° 15'.6 ; find the remaining 
parts and the areas of the two triangles which may be formed. 

2. Express the area of a parallelogram of which two adjoining 
sides are of given length, a and h, and make with each other a 
given angle 6\ 

3. Express the area of a triangle in terms of a base, c, and the 

two adjacent angles, a and /?. . ± a sin a sin f3 

•t*ns. <2,c —. t j -pjr. 
sin {a -f- /3) 

4. In a parallelogram is given a diagonal of length d, and the 
angles and cp which the diagonal makes with the two sides 
adjoining it. Express the area of the parallelogram. 

5. Express the area of the parallelogram in terms of the lengths, 
d and d' , of its diagonals, and the angle s at which they intersect. 

6. In a quadrilateral are given the four sides, a = 25.63, 
h = 24.09, c = 9.92, d = 29.97, and the angle, 78° 25 r , which 
the sides a and h form with each other. Compute the angles and 
the area of the quadrilateral. 



74 PLANE TRIGONOMETRY. 

7. A triangle AB C is to be divided into two parts of equal area 
by a line parallel to AB. What will be the ratios of the segments 
into which the other two sides are divided ? Ans. Vi : 1 — V'-J- 

8. A city lot fronts 60 feet on a street, and the parallel sides 
ran back, the one 100 and the other 135 feet. It is to be divided 
into two equal parts by a line parallel to its sides. What will be 
the frontage of each part, and the length of the dividing line? 

BemarJc. — The figure of the lot is a trapezoid, and the problem 
is that of dividing a trapezoid into two equal portions by a line 
parallel to the base. Let us put 

a, b, the parallel sides ; 

m : 1 — m, the ratio in which the non parallel sides, and there- 
fore the altitude, is divided by the dividing line ; 

k, the length of the dividing line. 

The unknown quantities of the problem are then m and Jc. If 
we put h for the altitude of the entire trapezoid, the altitudes of 
the two parts will be mh and (1 — m) h respectively. Therefore 
the areas will be . 7 7 . -, 

— ^— mh and — 1 —~ (1 — m ) h 

the equality of which gives the first condition. For the second 
condition we have the geometrical theorem that the difference be- 
tween the dividing line and either of the parallel sides is propor- 
tional to its distance from such side. This gives the proportion 

a — k : k — b :: m : 1 — m, 
whence (Jc — b) m = (a — Jc) (1 — m). 

The quotient of this equation by the preceding one gives an 
equation from which m is eliminated, and from which we find the 
value of Jc. a/^+E 

We then find for m the equation 

a — Jc 

m = 7. 

a — b 

Applying this method to the problem under consideration, we find 

Jc = 118.796 feet ; 
m — 0.462 97; 
frontages of lots, 27.778 and 32.222 feet. 



CHAPTEK VII. 

THE THEORY OF POLYGONS. 



63. A polygon is completely determined when the positions 
of its vertices, taken in regular order, are given. The polygon 
may then be formed by joining each pair of consecutive vertices 
by a straight line. 

The positions of the vertices may be defined by their co-ordi- 
nates, on a system now to be explained. 

64. Co-ordinates of a point. In geometry the position of a 
point is fixed by assigning to it certain lines or numbers indicating 
its situation relative to a fixed line, and a point on that line. 

Def. Any numbers or lines which determine the position of a 
point are called the co-ordinates of that point. 

Rectangular co-ordinates. Let OX be the fixed line of refer- 
ence, and O a point of reference on that line p 
from which we measure. 

Let P be a point whose position is to be 
expressed. From P drop a perpendicular 
PA upon OX. Then : 6 a X 

The line PA is called the ordinate of the point P. 

The line OA is called the abscissa of the point P. 

The ordinate and abscissa are called rectangular co-ordi- 
nates of the point. 

The indefinite line OX along which the abscissas are measured 
is called the axis of abscissas or the axis of X. 

The zero point from which the co-ordinates are measured is 
called the origin. 



76 PLANE TRIGONOMETRY. 

"When the rectangular co-ordinates are given the position of 
the point is completely determined. 

To find the point when -the ordinate and abscissa are given, we 
measure off from 0, on the line OX, a distance equal to the given 
abscissa. 

At the end of this distance we erect a perpendicular equal to 
the given ordinate. 

The end of this perpendicular will be the required position of 
the point. 

The abscissa of a point is represented by the symbol x, the 
ordinate by the symbol y. 

If x is positive, its length is laid off from toward the right ; 
if negative, toward the left 

If the ordinate y is positive, it is measured upward from the 
axis of X\ if negative, downward. 

Exercises. 

Draw a line OX 4 or 5 inches long as a line of reference, and 
lay off points having the following co-ordinates from a zero point 
near the middle of the line : 



1. 


x = 1, 


y = 


2i 


aches ; 




2. 


x = 2, 


y = 


1 


a 


3. 


x = 2, 


y = 


2 


« 


4. 


x = 1, 


y = 


1 


a 


5. 


X = 1, 


y = - 


■ 2 


a 


6. 


x = 2, 


y = - 


1 


a 


7. 


x — — 2, 


y = - 


• 2 


a 


8. 


x = — 1, 


y = 


2 


a 


9. 


x = 0, 


y = - 


■ 2 


u 


Polar co-ordmates. Dra\* 


r a line from to P, and call r i 


length and cp the 


angle XOP. 


Then 




P 


OD: 


= x = r cos < 


P, 




.-'' 




DP. 


= y = r sin < 


V, 








and Problem I. of Chapter V. 


reduces to : 


s^. . . , ... 





Given the co-ordinates x and y of a O D 

gomt, to find the distance and direction of the pomt from the origin. 



THEORY OF POLYGONS. 77 

Since the quantities r and cp completely determine the position 
of P, they are also co-ordinates. To distinguish them they are 
called polar co-ordinates. 

Exercise. 

Eight points have the following several co-ordinates. Find 
the values of cp and r (the values of r being all equal), and note 
■what relation exists among the values of cp. 



1. 


* = + 4, 


y = + 3; 


2. 


x = + 3, 


y = +4; 


3. 


x = — 3, 


y = - 4; 


4. 


a? = — 4, 


2/ = +3; 


5. 


a? = — 4, 


y = — «; 


6. 


a> = — 3, 


y = - 4; 


7. 


* = + 3, 


2/ = - 4; 


8. 


x = +4, 


y =-3. 



65. Definition of direction of lines and angle between 
them. Two finite lines which do not meet are considered to form 
a certain angle with each other ; namely, the angle which would 
be formed if they were continued until they met, or if a line paral- 
lel to the one were drawn through any point of the other. 

Since at the point of crossing four angles are formed, we may, 
in the absence of any convention, regard either of these angles as 
that between the lines. But as opposite angles are equal, these 
angles only have two different values. 

Also, in the absence of a convention, we may regard any angle 
as either positive or negative. Hence to a given inclination of 
the lines may be assigned any one of four different values, which 
values are divisible into two supplementary pairs. 

Example. If two lines intersect at an angle of 85°, we may 
consider their inclination, or the angle which they form, to be 

either 85°, 
or 95°, 
or - 85° = 275°, 
or - 95° = 265°. 



78 



PLANE TBIGONOMETRY. 



This ambiguity is avoided by the following conventions : 

1. We assign to each line a positive and a negative direction. 
The positive direction is that from the beginning to the end of a 
finite line. The angle they form is then that between their posi- 
tive directions. This is the same as the angle between two lines 
going out from the same point in the respective positive directions 
of the lines. 

2. We consider one side of the angle as that from which we 
measure, and we measure the angle to the other side in a positive 
direction, as explained in §§ 3 and 8. 

The four values are thus reduced to one. 

If two lines are parallel, their angle is 0° or 180° according as. 
they are similarly directed or oppositely directed. 



Projections of Lines. 

66, Def. The projection of a finite straight line AB upon 
an indefinite line XY is the distance A'B' between the feet of 
the perpendiculars from A and B upon the indefinite line. 
Bk b 





B^ 



To find the length of a projection. Through one end of the 

line, as A, draw A C parallel to XY, meeting BYm 0. Then 

AO=XY; 

AO=AB cos BAG. 
Hence 

Projection XY = AB cos BA C. (1) 

That is, 

The projection is equal to the length of the line projected into 
the cosine of the angle which the two Imesform with each other. 



THEORY OF POLYGONS. 



79 



Algebraic sign of a projection. Let the positive direction of 
the line OB be from to B, and let the line turn on O into the 
successive positions 00 and OD. 

If on the line of projection we re- 
gard directions toward the right as posi- 
tive, the projection A'B' will be positive. 

The whole line 00 will be projected 
at the point O; the projection will 
therefore be zero, a result given bj the 
formula (1), because cos 90° = 0. 

The projection A ' D' of OD will be negative because it falls 
in the negative direction. This also corresponds to the formula, 
because the angle between the two directions OD and A ' B' is 
obtuse. 

If we suppose OB to perform a complete revolution around O, 
we readily see that its projection goes through the same series of 
changes as the cosine of the angle which it forms with the line of 
projection. 




6*7. Projection of sides of a polygon. Let ABODE be any 
})olygon the positive directions of whose sides correspond with 
the circuit we should form in going round the polygon, so as 
to reach its vertices in alphabetical 
order. We shall then have, for the 
projections on the line X, 

Proj. of AB - A'B', which is + 

" BC=B'C, " - 

CD = CD', " - 

" DE=D'E', " — 

" EA = E'A, " +. 

The positive direction being arbitrary, we might equally take 
the directions AE, ED, DO, etc., as positive. Each of the pro- 
jections would then have the opposite algebraic sign from that 
just given. 




80 PLANE TBIGONOMETBY. 

The student will remark that the projection of the line is 
positive or negative according as the projection of its end is on 
the positive or negative side of the projection of its beginning. 
We wish now to determine the sum of the projections, and for 
this purpose must understand the algebraic addition of lines. 

68. Algebraic addition of lines. In geometry the sum of 

two segments AB and BCis denned , , , , 

as the segment A C, formed by put- A 

ting AB and BO end to end in the same straight line. 

In trigonometry and modern geometry we distinguish between 
the beginning and the end of each segment, and between the posi- 
tive and negative directions upon the segment ; the positive direc- 
tion being from the beginning toward the end ; the negative, 
from the end toward the beginning. 

When this distinction is attended to we must, in designating a 
segment by letters at its termini, write that letter first which is at 
the beginning of the segment, so that the letters shall follow each 
other in the positive order. The segment BA will then be the 
negative of the segment AB. 

We now generalize the definition of the addition of lines as 
follows : 

Def. The algebraic sum of several lines is formed by placing 
the beginning of each line after the first at the end of the line 
next preceding. This sum is then the segment from the begin- 
ning of the first line to the end of the last one. 

Example. — In the preceding figure we have 

AO+ CE=AE, 

as in geometry, because both segments are positive. 

But if we consider the segment CD as beginning at C and 
ending at D, then, by the above definition, the algebraic sum 
of the segments AC and CD will be the segment AD, from 
the beginning of A C to the end of CD. That is, a negative 
segment will be subtractive in the same way that in algebra 
the addition of a negative to a positive quantity implies sub- 
traction. 



THEORY OF POLYGONS. 



81 



In general, whenever A, B, C, D, E, ^represent points npon 
a straight line, we have 

AB + BO + CD + DE + EF = AE, 
however these points may be situated. 

If the end of the last line coincides with the beginning of the 
first, the sum will be zero, by definition. Hence, however the 
points A, B, and O may be situated, we have 

AB + BA = 0; 
AB + BO+ CA = 0. 

69. Let us now return to the projected polygon. On the pre- 
ceding system, the sum of the projections of the several sides upon 
the line X is 

A'B' + B'C / + CD' + D'E' + E'A' = 0. 

The same thing would be true if we took any other straight 
line as the line of projection. Hence : 

Theorem I. The algebraic sum of the projections of the sides 
of a polygon upon any straight line is zero. 

Since the projection of each side is equal to its length into the 
cosine of the angle it makes with the line of projection, this theo- 
rem may be expressed in the following form : 

If the sides of a polygon be a, b, c, etc., and the angles which 

these sides make with any straight line be a, /?, y y etc., we shall 

have 

a cos a -\- o cos p -f- c cos y -f- etc. = 0. 

We may imagine the sides of the polygon all taken up and 
placed with their beginnings at the 
same point, their length and direction, 
remaining unchanged. 

Their several projections will then 
have the same values as before, and 
in consequence the algebraic sum of 
the projections will still be zero. 

70. We now have the following 
theorem, the demonstration of which is left as an exercise for the 

student : 




82 



PLANE TBIGONOMETRY. 




Theorem II. If the algebraic sums of the projections of three 
or more straight lines upon any two non-parallel lines are each 
zero, these lines when placed end to end without changing their 
directions will form a polygon, the end of the last line falling 
upon the beginning of the first. 

Note. To fix the ideas the student may 
suppose the lines as first given to all ema- 
nate from one point, as in the last figure. 

The demonstration is begun by show- 
ing that in case the sum of the projections 
upon a straight line is zero, then, when the 
lines are placed end to end, the end of the 
last line and the beginning of the first must 
lie on the same perpendicular to the line of 
projection. Thus, in the figure, the sum of the projections of the four un- 
broken lines is zero, although they do not form a polygon. But, with such a 
figure, the projections will not be zero on any other non-parallel line. 

71. Theorem III. If a, b, c, etc., be the sides of a polygon, 
and a, j3, y, etc., the angles which these sides form with any 
straight line, we shall have 

a sin a -f- b sin /3 -|- c sin y -f- etc. = 0. 

Proof. Let OX be the base line from which the angles a, /?, 
y, etc., are counted ; AB, any side 
of the polygon; a, its length; 
a, the angle which AB makes 
with X 

Draw OY perpendicular to OX, 
and let PQ be the projection of 
AB npon OY. AB will then 
make with Y an angle 90° — a, and we shall have 
Projection PQ = a cos (90° — a) = a sin a. 

Treating all the other sides in the same way, the algebraic sum 
of their projections upon Y is found to be 

a sin a -\- b sin /? + c sin y -f- etc., 
which sum is zero by Theorem I. 

Theorem IY. If the sum of the projections of a series of 
straight lines upon any two non-parallel lines be zero, the sum of 
their projections upon any third line will be zero. 




THEORY OF POLYGONS. 83 

Note. This theorem follows immediately from Theorems I. and II., hut 
we prove it algebraically in order to show an elegant application of the addition 
theorem. 

Proof. Put 

ar, /?, y, etc., the angles which the straight lines make with one 
of the lines of projection ; 

7t, the angle which the first two lines of projection make with 
each other. 

Then a — jr, (3 — n, y — n, etc., will be the angles which the 
lines make with the second line of projection. 
By hypothesis we have 

a cos a -\- b cos /3 -f- c cos y -f- etc. = ; (a) 

a cos (a — 7t) -f- b cos (ft — it) -f- c cos (y — it) -f- etc. = 0. 
The last equation, by the addition theorem, reduces to 
cos n (a cos a -f- b cos f$ -\- c cos y -|~ . . . ) 
-|- sin ^ (a sin « + Z> sin J3 -\- c sin y -j- . . . ) = 0. 
The first term of this equation vanishes by (a). 
Hence, the whole sum being zero, the second term must also 
vanish, which requires that we either have 

sin n = 0, 
which will give 7t = or 180°, — in which case the two lines would 
be parallel, — or 

a sin a -\- b sin /3 -\- c sin y -f- etc. = 0. (b) 

Since, by hypothesis, the two lines are not parallel, the equa- 
tion b must hold true. 

Now let 6 be the angle which any third line of projection 
forms with the first line. The angles which the lines a, b, c, etc., 
form with this third line will then be 

a — 6, 0—0, y — 0, etc. 
Therefore the sum of the projections upon this line will be 

a cos (a — 6) -\- b cos (0 — 6) + c cos (y — 6) -f- etc., 
which reduces to 

cos 6 {a cos a -f- o cos /3 -\- e cos y + etc.) 
-\- sin 6 (a sin a -\- b sin y -f- c sin y -f- etc.), 
a sum which vanishes by (a) and (b), whatever be the value of 6 y 
thus proving the theorem. 



84 



PLANE TRIGONOMETRY. 



12. Cor. From §§ 69 and 71 it follows that if all the sides of 
a polygon but one are given in length and direction, — a, ft, c, etc., 
being the lengths, and ar, ft, y, etc., the angles with a fixed base 
line, — and if I be the omitted side and <? its angle, we shall have 
I sin S, = — a sin a — ft sin ft — c sin y — etc.; V . . 
I cos 8, = — a cos « — 5 cos y — c cos y — etc.; J 
which will determine I and <?, by Prob. I. Chap. Y. 

Example. A surveyor measures off courses and distances as 

follows : 

I. North 80° 28' east, 42.68 metres. 

II. North 23° 22' east, 22.79 " 

III. North 65° W west, 31.96 " 

IY. South 59° 58 r west, 40.13 " 

What distance and direction will 

carry him to his starting-point? 

"We note that the expression 

North r c east 

means 

r° east of north. 

Taking the east and west line 

OX as the base from which we 

o— — x 

measure angles, we readily find 

the angles made by each side with the base, as shown in the 

following table, which also gives the values of a sin a, ft sin yff, 

etc., and a cos «, ft cos ft, etc., as computed from the given data : 




Side. 


Length. 


Angle. 


a sin a, 
etc. 


a cos a, 
etc. 


a 
b 
c 
d 


42.68 
22.79 
31.96 
40.13 


9° 32' 

66° 38' 

155° 49' 

210° 2' 


+ 7.069 
+ 20.921 
+ 13.093 
- 20.085 


+ 42.090 
+ 9.039 

- 29.155 

- 34.742 




+ 20.998 


- 12.768 



Hence from (2) we have, for the last side, 
I sin Z = — 20.998; 
I cos 3 = + 12.768; 



THEORY OF POLYGONS. 85 

from which we find 

1 = 24.575; 

3 = 301° 18'.1. 

Expressed in the language of surveyors, the angle Z indicates the 

direction, South 31° 18'.2 East. 

It will be seen that we have taken as the positive direction of 
the last line that from the point last reached to the starting-point. 
This is in accordance with the convention that the positive direc- 
tions of the several sides of a polygon are so taken that in passing 
around it the directions shall all be positive or all negative. 

But we might equally consider the problem: Having pro- 
ceeded along a series of connected lines, AB, BO, etc., of which 
the lengths and directions are given, to E, what is our distance 
and direction from our starting-point A ? It is evident that the 
distance and direction are the length and direction of the line AE, 
which is simply the negative of the side EA necessary to com- 
plete the polygon. If we call e the angle of direction of AE, the 
equations for determining I and e would be . 

I sin e = a sin a -\- b sin /? -)- etc.; ) 
I cos s = a cos a -\- b cos /?, etc.; ) 
and, in the preceding example, we should have 
I sin s = -f 20.998 ; 

I cos e = — 12.768 ; 
which would give 

1 = 24.575; 

s = 121° 18'.1. 

Exercises. 

1. A surveyor ran a course S. 12° 13 r E., 289.26 metres, and 
thence 1ST. 82° 49 r E., 92.68 metres. What is his direction and 
distance from his starting-point ? "What is the direction and dis- 
tance of the starting-point from him ? 

2. Five sides of an irregular hexagon taken in order have the 
following lengths and directions relative to a fixed line : 

a. Length, 297.43 metres; direction, 332° 6 r .8 

b. " 606.07 " " 222°42 , .3 

c. " 421.02 " " 157° 59 / .8 



86 



PLANE TRIGONOMETRY. 




d. Length, 343.90 metres; direction, 5° 22'.1 ; 

e. " 40.92 " " 125° 2'.2. 
What is the length and direction of the remaining side \ 

73. Areas of polygons. When the sides of a polygon are 
all given in length and direction, the area may be computed by a 
process demonstrated in geometry, 
but which we shall describe here. 

Let ABCDE be any polygon, 
and OX the base line from which 
we measure angles. 

The area of this polygon is 

equal to 

Area A'ABCC 

c 
minus Area A 1 'AEDCC '. A' & ^T r/ c' ~ 

The first area is equal to the sum of the areas of the two 

trapezoids 

A' ABB' and B'BCC 

The second is equal to the sum of the areas of the three 
trapezoids g,^^ DIJ)EE >, and E'EAA'. 

It will be noted that there is one trapezoid for each side of the 
polygon, of which the non-parallel sides are the side of the poly- 
gon and its projection upon the base line. 

We have for the area of the first trapezoid, noting that the 
base line OX is perpendicular to the parallel sides, 

Area A' ABB' = \(AA' + BB f ) A'B'. 
Putting, as before, 

a, b, c, etc., for the length of the sides AB, BC, CD, etc.; 

a, /?, y, etc., for the angles which they form with OX; 

putting, also, 

_£>, the length AA', 
we have 

BB ' =p -f- a sin a ; 
A'B ' = a cos a. 
Substituting these values, we have, for the area of the first trapezoid, 
Area A' ABB' = i(2p -|" a sm a ) a cos a - 



THEORY OF POLYGONS. 87 

Passing on to the other trapezoids, we have, for the lengths of 
the several perpendiculars, 

BB ' = p -f- a sin a ; 
CC = p + a sin a -\- b sin /3 ; 
DD ' =p -\- a sin a-{-b sin ft -{-c sin y\ 
EE' — p -f- a sin a -f- b sin /ff + c sin y -\- d sin 6 

= p-esine; (§71) 

Also, for the altitudes of the trapezoids between their parallel 

sides, 

A'B ' = a cos a ; 

B'C = b cos /?; 

(7 / i> / = ccosr; 
i> / ^ 7/ r=^Zcos tf; 

^'J.' = 6 COS £. 

We thus have, for the several areas, 
%(2p -f- a sin a) & cos a ; 
2 (2j? + 2a sin a -\- b sin /?) 5 cos /? ; 
ii^P + 2a sm ** + 25 sin /? -|- c sm K) <? cos ^ ; 
2 (^p + 2a sin « -f - 25 sm ^ + 2c sin y -J- a 7 sin 6") a 7 cos 6" ; 
i(^P + 2a sin or -|- 25 sin /? -f- 2c sin y 

-\- 2d sin 6 -{- sin f) c cos £. 

It has been shown that the required area of the polygon is 
found by subtracting the last three of these areas from the first two. 

But in reaching this conclusion we took no account of algebraic 
signs, and so virtually considered the areas all positive. 'Now, as 
the figure is drawn, cos a and cos ft are positive, and the cosines 
of y, tf, and s are negative. Hence if we put the sign -|- before 
each of the areas (a), the subtraction will be indicated by the 
negative character of those products which have cos y, cos #, and 
cos s as factors, and so the algebraic expression will be correct. 

If we add up the quantities (a), beginning with the terms in p, 
we see that the coefficient of p is 

a cos a + b cos ft -\- . . . + e cos *> 
which is zero. Hence the quantity p disappears from the expres- 
sion for the area. Since p is defined as the distance below A at 



(a) 



PLANE TRIGONOMETRY. 



which the base line 0Xis drawn, this is the same as saying that 
the area of the polygon is independent of this distance, which evi- 
dently must be true. In fact, if we suppose the base line OX to 
move up or down, remaining parallel to itself, it will add and sub- 
tract equal areas to or from the positive and negative trapezoids, 
and so will leave the algebraic sum of the areas unchanged. 

Now let us suppose^? zero, and put 
y x = a sin a ; 

y 2 = %a sin a -f- 5 sin /3 = y x -|- a sin a -\- b sin /3 ; 
y 3 = 2a sin a ■+ 2h sin fi -J- c sin y = y 2 -+ I sin ft -f- c sin y ; t(5) 
^ = 2/ 3 + csin;r + ^sintf; 
y b = y 4 -|- ^ sin d -|- 6 sin f . 

(We remark that y 5 will come out equal to — e sin e if every- 
thing is correct.) 

We shall then have, for the double of the area of the polygon, 

2 Area = y^a cos a -+ yjb cos /? -+• y 3 e cos y -f- y 4 <# cos d -j- y 5 6 cos f. (4) 

As an example, let us compute the area of the polygon investi- 
gated in the example of §72. The following table shows the 
principal parts of the computation : 



a sin a, 

b sin j3, 

etc. 


Smns of Pairs. 


Vi, 2/2. 
etc. 


a COS a, 

6 cos /3, 
etc. 


Products. 




h 7.069 

- 20.921 

- 13.093 

- 20.085 

- 20.998 


+ 27.990 
+ 34.014 

- 6.992 

- 41.083 




- 7.069 

- 35.059 

- 69.073 

- 62.081 

- 20.998 


+ 42.090 
+ 9.039 

- 29.155 

- 34.742 
+ 12.768 


+ 297.5 
+ 316.9 

- 2013.9 

- 2156.8 
+ 268.1 










- 3288.2 



The first column gives the values of a sin a, ~b sin /3, c sin y> 
etc., already computed in the preceding example. 

The second column gives the values of a sin a -\- h sin /?, 
h sin j3 +- g sin y, etc., which are added to each value of y to form 
the value of y next following, as shown in (b). 

The third column gives the values of y„ y„ y 3 , y 4 , y b , computed 
by the formulae (5), from the numbers in the first two columns. 



THEORY OF POLYGONS. &g 

The fourth column gives the values of a cos a, b cos /?, c cos y, 
etc., already computed. 

The fifth column gives the products which enter into the 
equation (c). 

The algebraic sum being twice the area, we have 
Area of polygon = 1644.1 square metres. 

74. It will be noticed that the area of the polygon comes out 
negative, and the question arises, What interpretation is to be put 
on this result ? The answer is that the conventions of positive and 
negative as employed in this chapter, as applied to lines, express 
only the directions in which the lines are reckoned. By a change 
of direction the algebraic signs of all the trapezoids, and therefore 
of the area of the polygon itself, will be changed. A little con- 
sideration will show that this area will come out positive when we 
go round the polygon in what we have called the negative direc- 
tion, and vice versa. 

The algebraic sum of the areas, whether positive or negative, 
will always be the true area of the trapezoid under one important 
condition : that none of the sides cross each other. In this case the 
system of applying the algebraic signs will lead to the areas on the 
two sides of the point of crossing having opposite signs. Hence 
the result finally obtained will be the difference of the two areas. 

Exercises. 

1. If the lengths and directions of three of the four sides of a 
quadrilateral are 

a = 262.72 metres; a = 39° 49' 
I = 109.79 " /? = 150° 26' 

c = 300.63 " y = 242° 52' 

find the remaining side and the area. 

2. The sides of a quadrilateral, taken in regular order, have the 
following lengths and directions, in part : 

a = 29; a = 12° 26' 

I = 52; fi = 75° 58' 

c to be found; y = 172° 3' 

d = 66; d to be found. 
Find c and S. 




90 PLANE TBIGONOMETRY. 

3. Prove geometrically that if the sides of a polygon be joined 
together in any order, their directions 
remaining unaltered, the end of the 
last side will still fall upon the begin- 
ning of the first. 

An example of the construction is shown 
in the figure, where the sides are changed from' 
the order abed to adbc. 

4c. Express the lengths (x and a/) 
and the directions (6 and d') of the diagonals of a quadrilateral of 
which the lengths of the sides, taken in order, are a, ft, c, d, and 
their directions a, /?, y, 6\ 

It is only necessary to express the values of the quantities x sin 0, x cos 0, 
x' sin S', and x' cos 6' in terms of a, b, c, d, a, /3, etc. "We may suppose x and 6 
to refer to the diagonal from the beginning of a to the beginning of c, and x' 
and 6' to run from the end of a to the end of c. 

5. Using the same notation as in § 71, prove 

ft sin (j3 -— a) -\- e sin (y — a) -f- d sin (d — a) -\- etc. = ; 
a + ft cos (ft — a) -j- c cos (y — a) -j- d cos (S — a) -f- etc. = 0. 

6. If or, /?, and y are the angles which the sides a, ft, and c of 
a triangle make with a base line, and A 9 B, and (7 are the interior 
angles of the triangle, it is required — 

(1) To show 

A = 180° + fi — y ; B = 180° + y-a; C=180°+a-/3. 

(2) By combining the equations of Ex. 5, to deduce the law 
of sines (§ 56, 3) and the fundamental equations (§ 58, 9). 

7. From the same point emanate three lines, OA, OB, 00, 
of such lengths and directions that the sum of their projections 
upon any third line vanishes. If we complete the three parallelo- 
grams, of each of which two adjacent sides are two of the lines, 
the areas of these three parallelograms will be equal. 

Both a geometric and an algebraic proof may be given ; the former from 
§ 70, Th. I., the latter from § 62. 

8. From the corners A and B of a pentagonal field ABODE 
an engineer measures angles as follows : 

Angle BAG = 79° 23 r .6 ; Angle ABO = 47° 29 r .7 ; 

Angle BAB = 130° 7'.0 ; Angle ABB = 153° 42'.7 ; 

Angle BAE = 152° 40 / .2. Angle ABE = 164° 0'.8. 

AB measures 192 metres. Find the remaining sides and the area. 



CHAPTER VII. 

TRIGONOMETRIC DEVELOPMENTS* 



75. Lemma. When an arc becomes indefinitely small, the 
ratio of the sine to its a/rc approaches unity as its limit. 

Remark. In this lemma it is supposed that the arc is expressed 
in terms of radius as unity. 

Proof. It is laid down as an axiom of elementary geometry 
that when the number of sides of an inscribed 
regular polygon is indefinitely increased, the pe- 
rimeter of the polygon approaches the circum- ^' 
ference of the circle as its limit. That is, the <C x 
ratio x ^ x 

perimeter of polygon . . 

- - c , . ., approaches 1 indefinitely. 

circumference of circle rr J 

Kow, if we divide the circumference into n equal arcs, the n 
chords of these arcs, forming the perimeter of the inscribed poly- 
gon, will each be twice the sine of half the arc (§ 18). That is, 

we shall have 

Perimeter of polygon = 2n X sine of each arc ; 
Circumference of circle = 2n equal arcs. 
Therefore the above ratio is equal to 

sine of arc 
arc itself ' 

* The study of this chapter requires a knowledge of so much of series and im- 
aginary quantities as is contained in Book XL, Chapters L, II. , and V., and Book 
XII. , Chapter L, of the author's " College Algebra." It may be advantageously 
Tead in connection with Book XII. , Chapter II. , of that work. Students taking a 
partial course may pass to spherical trigonometry without reading this chapter. 



92 PLANE TRIGONOMETRY. 

which therefore approaches unity as its limit when n is increased 
indefinitely. 

76, Problem. To develop the sine and cosme of an angle m 
terms of the ascending powers of the angle. 
Let us suppose 

sin x = s + Sjpn + sji? -f- s z x z -\- s 4 cc 4 + s h x* -f- etc.; (1) 

cos x = c + c x x + o^ + G * x * + ^ 4 + CfP? + etc. (2) 

in which s , s„ 8 2 , etc., and o , e c„ etc., are coefficients whose values 

are to be determined from some known properties of the sine 

and cosine. 

Since sine (— x) == — since, the series for sin a? must change 
its algebraic sign when the algebraic sign of x changes. But only 
the odd powers of x will then change their sign. Therefore the 
even powers must not enter into the development, and we must 
have s = s 2 = s 4 = etc. = 0. 

The complete analytic proof of this proposition may be put into the following 
form. Changing x to — x in the development (1), we have 

sin (— x) = s — SiX-\- s 2 # 2 — s z x z -\- s^x* — etc. 
But we have, by changing the signs of both members of (1), 

— sin x — —s — s x x — SiX 2 — s 3 x z — SiX 4 — etc. 
Because these developments must be identically equal, we must have 

So — — So, 
*a = — «a, 
«4 = — S if 

etc. etc., 

which gives s = s» = s 4 = etc. = 0. 

Therefore the development of the sine becomes 

sin x = s x x + s 3 a? 3 + s b x b -f- etc. 

Dividing this equation by x, we have 

since . , 

— =s, + s 3 x* + s b x* + etc. 

Now suppose x to approach indefinitely near to zero. The 
first member will then, according to the lemma, approach unity as 
its limit, and the second member will approach s 1 as its limit. 
Therefore we must have s 1 = 1, and the development becomes 

sin x = x + s 3 x* -f- SfP? + etc. (3) 

Next take the development (2) for cos a?. If we suppose x = 0, 



TRIGONOMETRIC DEVELOPMENTS. 93 

we have cos x = 1. Therefore, putting x = 0, equation (2) will 
become 

1 = <?., 

which is the required value of c . 

Again, because cos (— x) = cos x, the development of cos x 
must remain unaltered when we change x into — x. 

Because this change will reverse the signs of all the odd powers 
of a?, the coefficients of o x , c„ etc., of these powers must all vanish, 
and the development must be 

cos x = 1 + <? 3 # 2 + c A x* -f- C 6 X* ... (4) 

77. We must now choose some property of the sine and 
cosine which will enable us to form equations of condition for the 
coefficients s 3 > s b ,etc, and c % , <? 4 , c t , etc. The most simple prop- 
erty for this purpose is that expressed by the addition theorem : 

sin (x-\-y) = sin x cos y -\- cos x sin y. (5) 

Because the equation (3) is to be true for all values of x, it 
must remain true when x -f- y is substituted for x. Making this 
substitution in (3), we have 

sin(a>+y) = x + y -f s 3 (x + yf+ s b (x + y) 5 + etc. 
Developing the powers in the second member, and collecting 
the terms multiplied by the first power of y, we have 

sin (x -\- y) = x -\- s 3 x 3 -f- s b x 5 -f- s,x 7 -f- . . . 1 

+ 2/(1 + 3^ 2 + 5^ + 7s,x 6 + ...)[ (6) 

-f- terms X y 2 , y\ etc., J 

which we need not compute. 

From (5) we have, by substituting for cos y and sin y their 
assumed developments, (3) and (4), 

sin (x+y) = sin x(l + ctf + ctf + ....) 1 

+ cos x(y + s 3 f + etc.) I (7) 

= sin x -\- y cos x -\- terms X y*, y% etc. J 
Now the expressions (6) and (7) must be identically equal; 
therefore the coefficients of each power of y must be identically 
equal. Equating the coefficients of the first power of y, we have 

1 -f- Ss 3 x 2 -f- 5s 5 a? 4 + 7s,x 6 -\- etc. = cos x. 
But we have, by (4), 

cos a? = 1 + o^ 2 + e 4 x* -f- etc. 



94 PLANE TRIGONOMETRY. 

This equation must be satisfied for all values of x. Equat- 
ing the coefficients of like powers of x, we find 

£:> w 

etc. etc.. 

Next consider the addition theorem for the cosine : 

cos (x -f- y) = cos x cos y — sin x sin y. (9) 

We find, by substituting x -\- y f or x in (4), 

cos (a> + y) = 1 + 0,(2! + y) 2 + cSx + 2/) 4 + etc.; 

from which, developing to the first power of y as before, 

cos (x -}- y) =1 + tf s a? a + <v# 4 + c 6 cc 6 -f- etc. ^ 

+ y(2<v» + 4^ + 6c 6 x b + etc.) L (10) 

+ terms X y% y\ etc. J 

From the second member of (9), by substituting for sin y and 

cos y their developments, namely, 

cos y = 1 -f- #y + etc., 

sin y = y + s 3 y 3 + etc., 
we find 

cos (x-\-y) = cos x-\-y(— sin a?) -f- terms X y 2 , y% etc. (11) 

Equating the coefficients of y in (10) and (11), we have 

2cje + 4c 4 a? 3 + 6c 6 x 5 + etc. = — sin x = — x — s 3 x* — s b x b — etc. 

Equating the coefficients of like powers of a?, 

a*=-l; 

4c 4 = - s s ; 

fo 6 = — *•- 
etc. etc. 

The equations (8) and (12), taken alternately, solve our problem. 

1 

~~2' 

2l 1 

3 



(12) 



From 


(12),, 


c 2 


a 


(8)„ 


*s 


u 


(12)„ 


o. 


a 


(8)« 


* s 







2. 


3 


> 


^3 


= 




1 




4 


2. 


3. 


4' 






1 







3s 

5 "" 2.3.4.5 



TRIGONOMETRIC DEVELOPMENTS. 95 

s t 1 

From (12)., 0. = - = - 2707576 ; 

etc. etc. etc. 

The law of the coefficients is obvious. ^ Substituting them in 

the developments (3) and (4), we have 



or , X° X' 

sin x = x — g-, + £- -j — ^ + etc. ; 



cos a? 



2! + 4!-6! + etC - 



(13) 



78. Convergence of the series. These series are convergent 
for all values of a?, a result which may be shown thus : 

The ratio of the successive pairs of adjacent terms in the 
development of the sine are, omitting the minus sign, 



X* 


X* 


x 2 


a? a 


2.3' 


4.5' 


6.7' 


8.9 7 



etc.; 

that is, each term is formed from the preceding one by multiply- 
ing by one of these factors. Now, however great may be x, we 
can continue these fractions so far that their denominators shall 
become greater than 2a? 2 , and so their values less than \. After 
this point the sum of all the following terms will be less than that 
of a geometric progression of which the first term is the term of 
the development (13), whose quotient by the preceding term is 
less than £, and whose ratio is \. Such a progression has a limit, 
whence the sum of the series (13) must also have a limit. 

The following two applications of these series will be useful 
as exercises : 

1. Square each series, carrying the square as far as the sixth 
or eighth power of a?, and show that the squares fulfil the condition 

sin 2 x -{- cos 2 x = 1 identically. 

2. Compute the values of the sine and cosine of 10° and 30° 
to 5 places of decimals, remembering that we are to take the 
natural measure of the arc x in radians (§ 14), and compare the 
result with that in the tables. 

We find, from § 14, 

Arc 10° = 0.174 53 ; 
Arc 30° = 0.523 60. 



PLANE TRIGONOMETRY. 



19. Sme and eosme m terms of imaginary exponentials. 
It is shown in algebra that if we call e the sum the series, 



1+1+ 

or 

we shall have 



2!~^~ 3!"t"4l 



e 



-f- etc., ad infinitum, 

828 .... , 



X X X 

^ = l + „, + r +g T + ri + eto. 



etc 



)••; 



Putting i, the imaginary unit, = V— 1, substituting xi = x 

for a?, and reducing, this equation becomes 

a? 2 , x* , f x 3 , x h 

^=l-2j + iy - etc. + ^-35 + 51 

or, from the developments (13), 

e** = cos x -f- i sin a?. 
Changing xi into — a^, we have 

e~ xi = cos x — i sin a?. 
The sum and difference of these equations are 

2 cos x = e** + e-** = e^ + -^ ; 

1 



(14) 
(140 



2* sin x — e* 1 — e' 00 * = e** — 



or 2 sin x = - (e** — e"**) 



e**' 



(15) 



= — i(e xi — e- xf ). 
For some purposes these equations may be written in the 
symmetric form 

e** -f- e-** = cos x + cos (— a>) ; ) 
#"* — e ~ "* = i sin x — * sin (— a?). S ^ ' 

These are two of the most celebrated equations in algebraic 
trigonometry, and are called Euler's equations, after their dis- 
coverer, Leonhaed Eulee. 

80. Demoivrds theorem. If in the equation (14) we substi- 
tute nx for x it becomes 

e™* = cos nx -f- i sin nx. 
By raising (14) to the nth power we have 

e nxi _ ( cos x -\-ism x) n . 



TRIGONOMETRIC DEVELOPMENTS. 



97 
(16) 



Therefore (cos x + i sin x) n = cos nx + i sin nx, 
which is known as Dcmoivre's theorem. 

This theorem enables us to develop the sines and cosines of 
multiples of an angle in powers of the sine or cosine of the simple 
angle, as follows : 

81. Problem. To develop sin nx and cos nx in powers of 
sin x and cos x. 

Developing the first member of (16) by the binomial theorem, 
and substituting for the powers of i their values (Algebra, § 325), 
namely, 

.'=-1, 

i' == —i, 
? = + l, 
etc. etc., 
we have * 

(cos x + i sin x) n == cos n x + W J 



-e) 



S)« 



cos n "" 3 x sin 3 x 



+ 



\j) cos" -4 x sin 4 # -f- etc. 



(17) 



This development being identically equal to the second mem- 
ber of (16), we have, by equating the real terms and putting, for 

brevity, 

c = cos x, s = sin x, 

cos nx = c n — [~j c n ~ 2 s 2 + (j) c n -V — (|) c n ' 6 s 6 + etc. 
This series will go on to infinity unless n is a positive integer, 



* We here use the very convenient abbreviated notation for the binomial co- 
efficients, namely: 



t) = t = " : 

We then have 



(^ \ _ n{n — 1) ( n — 2) 
3/ ~ 1.2.3 ; 

V s / ~" \n — sl 



1. 2.3. . . 8 



98 



PLANE TRIGONOMETRY. 



in which case it will terminate with a coefficient (— ], in which 

if 
j — n when n is even, and j = n — 1 when n is odd. 

If we suppose n equal to 2, 3, 4, etc., in succession, we have 

cos 2a? = c 2 — s 2 ; 
cos 3a? = c 3 — 3cs* ; 

cos 4a? = c 4 — zr 1 -^ cV 4- s 4 5 

1 . -a 



cos 5a? 



cos 6a? = c 6 — 



o" v +lr 



6.5 



+ 



6.5 



(18) 



1 . 2 ~ " '1.2 " 

We may make the results more uniform by substituting for 
the powers of s their values in powers of c, thus : 
s* = 1 - c 2 ; 
s 4 = i _ 2c 2 + c 4 ; 
s 6 = 1 - 3c 2 + 3 c 4 - c 6 ; 
etc. etc. 

Making these substitutions, and reducing the numerical coefficients, 

we find 

cos 2a? = 2c 2 — 1 ; 

cos 3a? = 4c 3 — 3c ; 

cos 4a? = 8c 4 — 8c 2 + 1 ; 
cos 5a? = 16c 5 - 20c 3 + 5c; 
cos 6x = 32c 6 - 48c 4 + 18c a - 1 ; J 
Next equating the coefficients of the imaginary terms of (17) 
to i sin nx and dividing by i, we find 

r n\ 

sin nx 



(19) 



|) C- 1 * - (|) c »-^+ (|) o— 5 * 5 - etc. 
= , j (j) c »-, _ (I) „-s , + g\ c „-s , _ et0 . f 



Supposing 7i = 2, 3, 4, 5, etc., this form gives 
sin 2a? — 2sc ; 
sin 3a? = s { 3c 2 — s a \ ; 

sin 4a? = s \ 4cc z — 4cs 2 } ; 
sin 5a? = s { 5c* — 10 c 2 s 2 -f- s 4 } ; 
sin 6a? = s \ Qc b - 20c* s* + 603'} ; 
etc. etc. 



(20) 



TRIGONOMETRIC DEVELOPMENTS. 99 

Substituting for the even powers of s their expressions (18), 
we find 

sin 3a? = s { 4c 2 — 1 \ ; 

sin 4a? = s {8c z — 4c J ; 

sin 5x = s jl6c 4 - 12c 2 + 1\ ; 

sin 6a? = * {32c 6 — 32c 3 + 6c\. 

Instead of substituting for the powers of s their expressions 
(18) in terms of the powers of c, we might have expressed the 
powers of c in terms of s, and by substituting them in (20) have 
developed the multiple sines in powers of s = sin a?. 

82. Expression /or powers of the cosine. The reverse prob- 
lem, to express the powers of the sine or cosine of an angle in 
terms of simple sines and cosines of multiples of the angle, is 
of yet more frequent application. 

Let us take the first equation (15), 

2 cos a? = e* 1 -f- c - **, 
and raise it to the wth power by the binomial theorem. We shall 
then have 

2 n cos n a? = e nxi + (j) e^~ 2)xi + (|) ^-«"«+ ... +«-■■* (21) 

To understand this general form let the student take the 
special cases n = 4 and n = 5. Then 

4 3 

2 4 cos 4 a? = e 4xi + 4c 2 ** + ^ -f 4c" 2 ** + e - to * ; 

l . w 

5 4 5 4 

2 5 cos 5 a? = c 5a * + 5c 3 ** + r-"— c** + -^ e-** + 5c -** + c" 5 **. 

Supposing n to be a positive integer, we see (1) that the co- 
efficients of terms equally distant from the two ends of the series are 
equal, and (2) that the exponents of e in such terms are equal and of 
opposite signs. Also, if n is even, the middle term does not contain 
a? ; but if n is odd, the terms on each side of the middle will con- 
tain a?. 

Therefore by joining the first and last terms, the term after 
the first and that before the last, etc., the development (21) may be 
put into the form 



(22) 



100 PLANE TRIGONOMETRY. 

2 n cos n a? = e^ + e~^+(-) { e (» -»>**_}- ^ --<»- a a* j 

Bnt by the general equations (15') we have, putting wa? for a?, 

^nori _|_ ^-nsi _ 2 CO g ^j — ; cos nx _|_ cog (_ n %^ 

whatever be the value of n. 

Hence, substituting this value of the exponential terms, 
2 n cos n a? = 

cos nx -f- [jj cos (^ — 2) x -\- ( x- J cos (n — 4) a?-f- etc. 

-f- cos (— nx) -f- (t ) cos (2 — 7^) x + ( ~ J cos (4 — w) a?-|- etc., 

the terms in the third line forming the end of the series, which 
is doubled over so that the end comes under the beginning. 
Giving to n the successive values 2, 3, 4, etc., we find 

2 2 cos 2 x = cos 2a? -f- 2 -f- cos (— 2a?) ; 

2 3 cos 3 x = cos 3a? -f- (f) cos a? -f- (J-) cos (— #)+(! ) cos (— 3a?) ; 

2 4 cos 4 x = cos 4a? -f- (f ) cos 2a? -f- ($) + (£) cos (— 2a?) 

+ a)cos(-4a3); 

2 5 cos 6 x = cos 5a? + (f-) cos 3a? -|- (f) cos a? 

+ (|) cos (- 5x) + (£) cos (- 3a?) + (f) cos (- a?). 
"We have extended the series in this form that the student may 
see the law of its formation, which is as follows : The successive 
coefficients are the binomial coefficients. The coefficient of a? in 
the first term is n, and it diminishes by 2 in each following term, 
so as to become — n in the last term. 
The two well-known formulae 

cos (— nx) = cos nx, 

'7/ = \n~^~s)> 
will enable us to combine every pair of terms equally distant from 
the extremes into a single one. For instance, we have 

U/ ~~ 1.2.3.4 ~ 1.2"" \2/ 5 



(23) 



/5\__ 5.4.3.2 /5\ 
\4/- 1.2.3.4 ~ \l) ~ 5 - 



TRIGONOMETRIC DEVELOPMENTS. 



101 



Combining them thus and dividing each equation by 2, we find 
2 cos 2 x = cos 2a* + 1 ; 
2 2 cos 3 x = cos 3a? -f- 3 cos x ; 

2 s cos 4 x = cos 4a? -J- 4 cos 2a? -|- 



1 4.3 



2* cos 5 a? = cos 5a? -f- 5 cos 3a? -f 



2 '1.2 
5.4 



ij— 2 cosa?; 



(24) 



2 5 cos 8 a? = cos 6x -f- 6 cos 4a? -f- — ^ cos 2a? -f- ^ . ' ' ; 
etc. etc. etc. 

A careful study of these forms will show the general law of 
formation, and enable us to carry the powers forward to any 
extent. 

83. Powers of the sine. To find the powers of the sine we 
take the second of equations (15), 

2i sin a? = e* — e ~ ■*, 
and raise it to the Tith power by the binomial theorem. "We then 
have 



nxi 



2 n i n sin n a? = 



_ ^^(» 2)xi _j_ (^ e (n-4)xl _J_ etc# 



{n-2)xi 



2J e 



+ etc. 



(25) 



The bottom line gives the last terms of the series, arranged in 
the reverse order, so that terms equally distant from the extremes 
are under each other. 

The upper signs are to be used in the bottom line when n is 
even • the lower ones when n is odd. This will be seen by form- 
ing the developments f or n = 5 and n = 6. 



2H 5 sin 5 a? = e 5 ** - (f) e te * + ($)«* - 
2 6 i 6 sin 6 a? 



-fxrt ^ 



e **i _ (|) ^+ (f) J* -\-e-^ 



(250 



"We first take the case of n even. The nth. power of i is then 
+ 1 or — 1 according as in is even or odd. Adding the terms 
of (25) and substituting for the exponential functions their values 
in terms of cosines from (15'), we find, when n is even, 



102 PLANE TRIGONOMETRY. 

w (n\ 

(—1)2 2 W sin n a? = cos nx — ( - J cos (n — 2)a? 



i 



(26) 



+ ( -J cos (n — 4:)% — etc. 

When n is odd, the sum of each pair of corresponding terms 
in (25) will give rise to a sine. Making the substitution and divid- 
ing by i, we shall have 

2H n ~ 1 sin n a? = sin nx — (r-J sin (n — 2)a? 



-f- ( a) sm ( n — 4)a? — et c. 



(26') 



Giving n the successive values 2, 3, 4, etc., and applying the 
forms (26) and (26') alternately, and changing all the signs in (26) 
when %n is odd, we find 

2 a sin 2 x = — 2 cos 2a? + 2. 

2 3 sin 3 x = — sin 3a? -f- 3 sin x — 3 sin (— a?) -f- sin (— 3a?), 
or 2 2 sin 3 a? = — sin 3a? -|- 3 sin a?. 

2 4 sin 4 a? = cos 4a? — (j-) cos 2a? -f- (|) — (f) cos (— 2a?) 

+ (f)cos (-to), 

or 2 8 sin* a? = cos 4a? — 4 cos 2a? + 3. 

2 6 sin 6 a? = sin 5a? — (■{■) sin 3a? -f- ■§• sin a? — sin (— 5a?) 
+ (J) sin (- 3a?) - (D sin (- a), 
or 2 4 sin 5 a? = sin 5a? — 5 sin 3a? -f- 10 sin a?. 

2 6 sin 6 x— — cos 6a? + (£) cos 4a? — (f ) cos 2a? + (f) 

— cos (6a?) + (| ) cos (— 4a?) — (f ) cos ( — 2a?), 

6.5.4 
or 2 6 sin 6 x = — cos 6a? + 6 cos 4a? — 15 cos 2a? -f- i . 

etc. etc. 



1.2.3' 



84. We might have obtained the above results for cos a?, 
cos a a?, cos 3 a?, etc., and sin a?, sin 2 a?, etc., by consecutive multiplica- 
tion, and substitution of sines or cosines of sums for products, by 
§ 43. Thus, by § 44, eq. 16', 

2 cos 8 a? = cos 2a? ~|- 1 ; 
X 2 cos a?, 

2 2 cos 3 a? = 2 cos a? cos 2a? -f- 2 cos a?. 



TRIGONOMETRIC DEVELOPMENTS. 103 

.Substituting for the first term of the second member its values 
(12) of § 43, we have 

2' cos 3 x = cos Sx -[- cos a? -\- 2 cos x = cos 3a? + 3 cos x. 
Multiplying this equation by 2 cos a?, we should have an ex- 
pression for cos 4 x, etc. But the use of exponentials enables us 
not only to obtain the higher powers more expeditiously, but to 
find the general law of the series, which is not readily done by 
multiplication. 

Exeecises. 

1. In the expression 

1 -f- 2 cos x + 3 cos 8 x -f- 4 cos 3 x, 
substitute for the powers of x their values in terms of the multiple 
of x, and reduce the expression to one containing simple multi- 
ples of x. 

Solution. From (24), 

4 cos 3 x = cos 3a? -{- S cos x 

3 cos 3 x = -f- f cos 2a? +f 

2 cos x = + 2 cos a? 

1 = 1 

Sum = cos 3a? -f- f cos 2a? + 5 cos a? + f 

2. Reduce the expressions 

4 cos 3 a? — 3 cos x, 
8 cos 4 x — 8 cos 2 a? -(- 1, 
16 cos 5 a? — 20 cos 3 a? -f- 5 cos a?, 
32 cos 6 a? — 48 cos 4 a? -|- 18 cos 2 x — 1, 
to terms containing sines and cosines of multiples of a?, thus prov- 
ing eq. (19). 

3. Prove that the expression 

1 — 2a cos 6 -f- a? 
may be resolved into the two factors (1 — ae 0i ) (1 — ae~ H ). 

4. Eesolve the expression x? n — 2x n cos 6 -\- 1 into the product 
of two factors, as in the last example. 



104 PLANE TRIGONOMETRY. 



Trigonometric Forms of Imaginary Expressions. 

85, It is shown in algebra that an imaginary or complex ex- 
pression may be reduced to a certain number of real units plus a 
certain number of imaginary units. If we put 
i, the imaginary unit, = V — 1, 

a, the number of real units, 

b, the number of imaginary units, 
the complex expression will be 

a + bi. (1) 

¥e have already shown (§ 47) that, whatever be the numbers 

a and b, we can find a positive number r and an angle cp y such that 

r cos cp = a; 
r sin <p = b. 
If we substitute these values of r and cp in (1) it will become 

a-\-bi = r (cos (p-\-i$m cp). 
But equation (14) gives 

cos cp -f- i sin cp = e*K 
Therefore a + bi = re* 1 . (2) 

We hence conclude : 

Every complex expression com be reduced to the form 

re* 1 , 
which is called the general form of the complex expression. 
The coefficient r is called the modulus of the expression. 
A yet better term, used by the Germans, is the " absolute value" 
of the expression. 

The angle cp is called the argument of the expression. 
Example. Reduce the expression 

- 0.9223 + 1.0962a 
to the general form. 

Putting r cos cp = — 0.9223, 

r sin cp = 1.0962, 
and applying the process of §47, we find 

r = 1.4326 ; 
cp = 130° 4'.54. 



TRIGONOMETRIC DEVELOPMENTS. 105 

This process being purely algebraic, the angle cp should be ex- 
pressed in radial units. Reducing to this unit, we find 

cp = 2.2703. 

Therefore the required general form is 

- 0.9223 + 1.0962* = 1.4326^™*. 

The student who is acquainted with the geometric represen- 
tation of imaginary quantities will see that the quantity r corre- 
sponds to the modulus and <p to the angle of the complex expres- 
sion as defined in algebra. 

The geometric construction of the expression a -f- hi is effected 
by laying off the length a on the axis of X, and at the end of this 
length erecting a perpendicular equal to b. 
If be the origin, we shall ha^e 
OX = a; 
XY=b. 
Then joining Y we shall have 
OY=r; 
Angle XOY = cp. 

86. Multiplication of complex expressions m the general 
form. If any two complex expressions are 

re* 1 and qe ei , 
we have by multiplying them 

roe® + W. 

This is another complex expression of the general form of 
which rq is the modulus and cp -|- 6 the argument. Hence : 

The modulus of a product is equal to the product of the 
moduli of the factors. 

The argument of a product is the sum of the arguments of the 
factors. 

If we multiply n equal factors, each represented by re**, the 
result will be 

Hence : 

The modulus of a power is equal to the corresponding power 
of the modulus of the root. 




106 PLANE TRIGONOMETRY. 

The argument of the power is the argument of the root multi- 
plied by the index of the power. 

8*7. Periodicity of the imaginary exponential. From the 
known equations (§ 24) 

cos (cp -\- 27t) = cos cp, 

sin (cp -f- 27t) = sin cp, 

and the following equations given by the preceding theory, 

e® + ^ = cos (cp -\-2rt)-{-i sin (cp + 2it), 
e <i>i __ cos ^ _|_ i s j n ^ 

we have 

re<& + 2T)i — . ^.^i . 

that is : 

The value of a complex quantity remains unaltered when we 
increase its argument by a circumference. 

Since the addition of one circumference does not change it, 
the addition of any number of circumferences will still leave it 
unchanged. Hence : 

If the argument of a complex quantity increases indefinitely, 
the values of the quantity itself will repeat themselves with every 
circumference by which the argument increases. 

A quantity whose value repeats itself in this way is said to be 
periodic. 

88. Let us next inquire for what special values of cp the ex- 
ponential function efi* will be equal to the real or imaginary unit. 
Considering again the equation 

e 4>i __ cos cp-\-i sin cp, 
we notice that sin cp = whenever cp is a multiple of 180° or of 
7t. When the multiple of it is even, we have cos cp = -j- 1; 
and when it is odd, cos cp = — 1. Hence, putting 
cp = n, 2#, Zn, etc., 
e~ wi = — 1 

e - 2ni — _|_ 1 



we have 

e m = — 1 



e 2m — _|_ 1 



Ziti 



(a) 



etc. etc. 

In order that cos cp may vanish, the angle cp must be 90°, 270°, 

450 o , etc. ; that is, it must be an odd multiple of \n. Sin cp will 



TMGOXOMETRIC DEVELOPMENTS. 107 

then be + 1 or — 1. Putting cp = \n, cp = |^, cp = \n, etc., 
on both sides of the preceding equation, we have 

e \iti = _|_ l . e - ini — _ i . -J 

€ 3« = _ {' e ~ ini = _|_ ^ I ^ 

e \ai _ _|_ ^ . ^ ~ 2^i — — £ j 

By squaring each of these equations we shall reproduce the alter- 
nate equations (a). 

89. Hoots of unit?/. The foregoing theory enables us to find 
very simple and elegant expressions for the roots of the equation 

ap — 1 = 0, 
or x n = 1. 

From the general theory of equations, the equation x n — 1 = 0, 
being of the nth degree, must have n roots ; that is, there are n 
quantities which, being raised to the nth power, will produce 1. 

These quantities are called the nth roots of unity. 

Because l n is always 1, whatever be n, -f- 1 is itself one of the 
nth. roots of unity. 

Because (— l) n = 1 when n is even, — 1 is always an nth root 
of unity when n is even. 

Hence one or two of the n roots of unity, viz. + 1 an d — 1> 
are real ; all the others are imaginary. 

90. Problem. To find the nth roots of unity. 

Solution. Let a required root be re ei , r and 6 being quantities 
to be determined. By the requirements of the problem, the nth 
power of this quantity must be 1. Its nth power is 

(re e€ ) n = r n e nBi = r n (cos nO -f- i sin n6). 
In order that this expression may be equal to unity, a real quan- 
tity, the coefficient of i must vanish, and we must have 

sin nO = 0, 
which gives 

cos nd = 1. 
Hence 

r n = 1, 

which is satisfied by supposing 

r = 1. 
We must also have 

nd = or 27r or 4^ or 67T, etc. 



108 



PLANE TRIGONOMETRY. 



Dividing by ft, we see that 6 may have any one of the values 

= 0, 



6 



2~, 



= 4-, 



6 



ft' 



etc. 
By substituting, in the assumed expression, re ei for the value of 
the root, we have 

2v in- 6w 

nth. roots of 1 = 1, e n , e n , e n , etc. 
Reducing to the trigonometric form, these expressions become 

i; 

cos 2- + i sin 2- : 
W ' ft ' 

cos 4 — L- ^ sin 4- ; 

ft ' ft ' 

cos 6 — h * sm 6- : 

ft ' ft ' 



etc. 



etc. 



The angle increases by 2- with each root, and by writing n con- 
secutive values we shall be carried all round the circle. 

The solution which we thus reach may be represented thus : 
Divide the circle into n equal arcs. 

Let the length of each arc he or, so that na = 360° = 2?r. 
The nth roots of unity will he : 

cos -f- i sin = 1 ; 

cos a -\- i sin a ; 

cos 2a -\- i sin 2a ; 

cos Ba -f- i sin Ba ; 



(3) 



cos (ft — l)a -f- i sin (ft -— l)a. 
Example. To find the sixth roots of 1. Here 



ft = 6; 



180° 



ft 



30 c 



2- = 60° = a. 
n 



TRIOONOMETEIC DEVELOPMENTS. 



109 



Hence the six roots are 
1: 



cos 60° + i sin 60 c 



1 VI . 

2 + 2 *' 

VI 



(4) 



cos 120° + i sin 120° = — ^ + -^ i ; 
cos 180° + * sin 180° = - 1 ; 

l Vs 

cos 240° + i sin 240° = - - _ — i ; 

1 V3 
cos 300° + i sin 300° = ^ — -y *. 

The result can be readily proved by raising each of these quan- 
tities to the sixth power. 

The roots may also be constructed as in the anexed figure. 

PA 

If angle XOA = AOB = etc. = 60°, then, since -y^ = 

OP 

sin 60° and -^ = cos 60°, 



1, 



OX 

ox~ 

OP PA 

ox + ox'' 

OQ , BQ 



h 





A 


/ 








/ 




/ 








\ / 




\ / 




y' 


P | 


A 




/0\ 




/ \ 




' \ 








\ 








\ 





ox + ox 

etc. etc., 
are the sixth roots of unity. 

EXEKCISES. 

1. Find and construct the eighth roots of unity, or the roots of 
the equation x* — 1 = 0. 

2. Find the roots of the equation x 12 — 1 = 0. 

91. Relations between the roots of unity. If we represent 

by x any such quantity as cos a -f- i sin or, we have, by what 

precedes, 

x 2 = cos 2a -\- i sin 2a ; 

a? 3 = cos 3a 4- i sin da; 



x" = cos na 



% sin na. 



110 PLANE TRIGONOMETRY. 

Hence the formation of the powers of x may he represented 
geometrically by laying off equal arcs around a circle. 

If x is any nth root of unity, then measuring off its angle a 
n times will bring ns back to the starting-point. 

If a is itself the nth. part of the circumference, then the re- 
maining roots as given in (3) are the first n powers of x. 
Hence : 

All the roots of unity are powers of the root corresponding to 
the smallest arc. 

From this it follows that if we measure off with a pair of di- 
viders, from to any division-point, the mth, for instance, and repeat 
the measure n times, the nth measure will end at the zero-point. 

This is evident of itself, because n measures of m arcs each will 
measure off mn arcs ; and because n of the arcs make up a circum- 
ference, the mn arcs will extend around the circle exactly m times. 

But it does not follow that any such series of n measures will 
include all the roots. Suppose, for example, that in the preceding 
figure, where n = 15, we measure arcs of 6a. The 15 successive 
points reached with the dividers will then be 
0, 6, 12, 3, 9, 0, 6, 12, 3, 9, 0, 6, 12, 3, 9, 0. 
This series includes only 5 of the points of 
division, each of these 5 being repeated 3 Ti 
times, while the remaining 10 have not been 
included at all. 

If we take the measure 4a in our divid- 
ers, the points of division included in the 
series will be 

0, 4, 8, 12, 1, 5, 9, 13, 2, 6, 10, 14, 3, 7, 11, 0, 
which comprise all n points. Hence in this case all the roots are 
powers of x* or of cos 4a -f- i sin 4a. 

92. We now have the following proposition, where we put 

360° 




a = 



n 



If m is prime to n, all the nth roots of unity may be repre- 
sented as powers of cos ma -f- i sin ma. 



TRIG0N0MET1UG DEVELOPMENTS. \\\ 

Proof. Starting from any point of the circle, and measuring 
off equal arcs, each of length ma, let p be the smallest number of 
measures which will bring us back to the starting-point. The 
total length of arc measured off will then bepma. 

Since we are brought back to the starting-point, we must have 
measured off an entire number of circumferences. Let q be that 
entire number. 

Because each circumference = na y the whole q circumferences 
measured = qna. Therefore 

pmoc = qna, 

and - = — . 

q m 

Because, by hypothesis, m and n are prime to each other, the 
fraction — is irreducible, and the smallest values of p and a are n 

and m respectively. 

Therefore any n measures will end at n different points of the 
circumference, and will therefore include all n points. 

Def. A root of unity whose powers include all other roots is 
called & primitive root. 

Cor. If n is a prime number, all the roots are primitive roots. 

Exercises. 

1. The 15th roots of unity being 

cos 24° + i sin 24°, cos 48° + i sin 48°, etc., 
it is required to find which of these roots are primitive. 
Prove the following propositions : 

2. If n is a prime number, all the wth roots of unity are prim- 
itive roots. 

3. If x be any primitive n\h root of unity, and if p be any 
number prime to n, then x p will also be a primitive root. 

Note. In the preceding theorem this is proved for the case when x is the 
root corresponding to the smallest angle. The proposition now enunciated ex- 
tends to the case in which we start from any multiple of this angle prime to n. 



PART II. 
SPHERICAL TRIGONOMETRY, 



CHAPTER I. 

FUNDAMEN1AL PRINCIPLES. 



93. Def. Spherical trigonometry treats of the relations 
among the six parts of a trihedral angle. 

Def. The six parts of a trihedral angle are its three face-angles 
and its three edge-angles. 

94. Representation of a trihedral angle by a spherical tri- 
angle. If be the vertex of a trihedral angle, and OM, OR, and 
OQ its three edges, we may con- 
struct a sphere having its centre at 
0, and having an arbitrary radius 
OA. The spherical surface will then 
cut the edges at the three points A, 
B, and equally distant from 0. 

The three faces OMR, ORQ, 
OQM will intersect the spherical 
surface in three arcs of great circles, AB, BO, OA, which arcs 
form a spherical triangle. 

It is shown in geometry that the three angles A, B, and G of 
the spherical triangle are equal to the respective edge-angles OM, 
OR, and OQ of the trihedral angle. It is also shown that the 
arcs AB, BO, and OA, which form the sides of the triangle, 
measure the respective face-angles MOR, ROQ, QOM of the 
trihedral angle. 




114 SPHERICAL TRIGONOMETRY. 

Therefore the six parts of the trihedral angle are represented 
by the corresponding parts of the spherical triangle, and the rela- 
tions among the parts of the one are the same as the relations 
among the parts of the other. 

The term spherical trigonometry is applied because the investi- 
gations are generally made by means of the spherical triangle. 

A trihedral angle, with its corresponding spherical triangle, may be readily- 
constructed as follows : Cut a circular disk of pasteboard or stiff paper, from 
four to six inches or more in diameter. From this disk cut out a sector of any 
magnitude. It will be well to have several disks with sectors ranging from 45° 
to 200° cut out. Divide the remainder of the disk by two radii into three 
sectors, such that the greatest shall be less than the sum of the other two. Bend 
the disk along each of the two dividing radii, cutting the latter part of the way 
through if necessary, and bring the extreme radii together. We shall then have 
a figure like O-ABG of the preceding diagram, the three plane sides forming the 
trihedral angle, and the three arcs bounding the edge of the disk forming the 
spherical triangle. 

95. General remarks upon spherical triangles. A spherical 
triangle may be defined as that figure which is formed by joining 
any three points on the surface of a sphere by arcs of great circles. 
The three points will then be the vertices of the triangle. 

But between any two points we may draw two arcs of a great 
circle, which together make up a complete great circle through 
the points. One of these arcs will be less, the other greater, than 
180°. To avoid ambiguity, the arc less than 180° is supposed 
to be taken, unless otherwise expressed. We therefore adopt the 
rule : 

Each side of a spherical triangle is supposed less than 180°, 
unless otherwise expressed. 

This rule is a mere convention, which may be set aside whenever we desire 
to give greater generality to our conclusions. Nothing prevents us from sup- 
posing ourselves to pass from one vertex to another by passing several times 
around the sphere. The corresponding side of the triangle will then consist of 
several coincident great circles plus either of the arcs joining the vertices. If 
we suppose a to be the shorter arc joining two vertices, the general arc measure 
of the side through those vertices will be 

n 360° + a or (n + 1) 360° - a. 

96. Every spherical triangle encloses a portion of the spherical 
surface, forming the area of the triangle. We then have the theorem : 




FUNDAMENTAL PBINCIPL, li;, 

Three great eirclea divide the surface of the sphere into eight 
triangular portions. 

This La shown as follows : One great circle 
divides the surface into two equal parts. A 
second great circle intersects the first in two 
points, and divides each of those parts into 
two lunes, so that the whole surface is then C'V~~ ^~"~/\y 

divided into four lunes. A third great circle *" -""" 

cuts through all four of these lunes, and forms T f ei f ht ^ he ™ a A tri - 

o ' angles formed by three 

eight spherical triangles. great circles. 

In the same way, any two planes divide the space around their 
line of intersection into four parts. A third plane intersecting 
them divides the space around their point of intersection into 
eight parts, forming eight trihedral angles. 

Remark. The student should guard himself against consider- 
ing a figure of which either side is a small circle of the sphere as 
a spherical triangle. For example, the figure formed by two arcs 
of meridians and a parallel of latitude is not a spherical triangle. 
Such figures do not represent the parts of a trihedral angle, and so 
do not correspond to the definition of a spherical triangle. All 
the important problems connected with them may be reduced to 
problems of spherical trigonometry, so that there is no need of 
giving them special consideration. 

Exercises. 
The following exercises are introduced to test the student's fundamental 
conceptions of spherical geometry, and especially of the relations of great circles 
of the sphere. Their successful performance will show that he is prepared to 
take up the subject of spherical trigonometry with advantage. A globe, on 
which figures may be drawn at pleasure, will be of great service in assisting his 
conceptions, and should be made use of whenever practicable. 

1. A and A' are two opposite points on a sphere. If any third 
point Xbe taken on the sphere, to what constant arc will the sum 
XA-\- XA' be equal, and what will be the angle AX A' ? 

Note. Opposite points are those at the ends of a diameter. 

2. If one side of a spherical triangle be equal to a semicircle, 
what relations will then subsist between the other two sides? 
What will be the magnitude of the opposite angle ? 



116 SPHERICAL TRIGONOMETRY. 

3. Let A, B, and O be the three'vertices of a spherical triangle ; 
a, b, and c, the sides opposite these vertices respectively ; A\ B\ 
and C\ the points opposite the vertices. It is then required : 

(a) To express by the letters at their vertices the eight tri- 
angles which will be formed when each side of the original 
triangle ABO is produced into a great circle. 

(b) To express the, sides of each of these eight triangles in 
terms of a, b, and c, making use of the theorem that any two great 
circles intersect each other in two opposite points. 

(c) To express the angles in terms of the angles of the original 
triangle, which we may represent by the letters A, B, and O 
marking their vertices. 

(d) It being found that the eight triangles are divisible into 
four pairs, such that the sides and angles of each pair are equal, it 
is required to show the relations of each pair. 

4. If one angle of a spherical triangle is A, show that the sum 
of the other two angles is contained between the limits 180° — A 
and 180° + A. 

Note. If the student finds any difficulty in this question he may begin by 
supposing the triangle to be isosceles, and the two equal sides to increase from 
0° to 180°. 

5. Hence show that the spherical excess cannot exceed twice 
the smallest angle. 

6. If the three sides of an equilateral spherical triangle be con- 
tinually and equally increased, what is the limit of their sum? 
What is the limit of the angles as the sum approaches its limit ? 

9*7. Fundamental equations. Let us put 
#, b, c, the three face-angles of the trihedral angle — that is, the 
angles subtended by the three sides of the spherical triangle; 

A, B, C, the opposite edge-angles of 
the trihedral angle, or the angles of the ^K^\\ 

spherical triangle. ^^"^ \\ \\ 

Then if O-ABC be any trihedral °<^-—~~ 
angle, we shall have N. 

a = angle BOC; \ 

b = angle CO A ; 
c = angle A OB. 



FUNDAMENTAL PRINCIPLES. 117 

Through any point A of OA pass a plane perpendicular to 
OA, and let B and C be the points in which it meets the other 
two edges. We shall then have 

A = angle BA C, 
while OAB and OA will both be right angles. 
In the triangle BOO we have 

BO 2 = OB 1 + OG 2 - 20B . 00 cos a. (§ 58) 

In the triangle BA we have 

BO 2 = AB 2 + A0 2 - 2^.5.^.^008^. 
Equating these two values of BO 2 , and transposing, we find 
20B . 00 cos a = OB 2 - AB 2 + 00 2 - A0 2 +2AB . A0 cos A. 
But, because OAB and ^J.C'are right angles, 
OB 2 - AB 2 = OA 2 ; 
0G- - AC 2 = OA 2 . 
Substituting these values, and dividing by 2 OB . 00, we have 
OA OA AB AC 
C0Sa =0B'0C + 0B'0C C0QA ' 
Now 0A 

OB = C0S ' ; 
OA 

oc =cosh > 

AB 
0B= smc ' 

oo= smk 

Therefore cos a — cos 5 cos c _j_ sm 5 sm c C08 ^ (#) 

By treating the other edges in order in the same way, we 
obtain two more equations, which may be written by simply per- 
muting a, b, and c and A, B, and O circularly ; that is, by substi- 
tuting for each letter the one next in order, a following c. Thus 
we have the system of three equations: 

cos a = cos b cos c + sin b sin c cos A ; i 
cos b = cos c cos a -f- sin <? sin & cos 5 ; V (1) 

cos c = cos a cos 5 -f- sin <z sin J cos (7. J 
These three equations are the fundamental equations of spherical 
trigonometry, because by means of them, when three parts are 



118 SPHEBICAL TRIGONOMETRY. 

given, the other three may be found. For practical application 
they are transformed and simplified in numerous ways. 

98. Permutation of parts. We have deduced the equations 
(1) from (a) by merely permuting the letters. This process 
may be applied generally in accordance with the following 
theorem : 

If we have found between the parts of a spherical triangle any 
equation which is true for all triangles, it will remain true when 
we permute the sides in any way ; provided that we also permute 
the opposite angles in the same way. 

For if we have proved our equation by calling the three 
sides a, b, and c, and the opposite angles A, B, and O, we could 
apply the same proof to the other parts of the triangle, substi- 
tuting 

Side a for side b, and vice versa, 
and 

Angle A for angle B, and vice versa, 

in the demonstration. We should then have a result in which a 
and b changed places, and A and B changed places. 

By interchanging a and c, and b and c, with their opposite 
angles in the same way, we should form all the six equations 
which could be written by permuting the symbols in the way 
described. 

If, however, we made any supposition respecting any side or 
angle such that the reasoning applied to it would not apply to the 
others, then the symbol of this side or angle could not be per- 
muted. For instance, we cannot permute all the parts in a formula 
true only for right triangles. 

It follows from this that any true formula which expresses the 
value of one part in terms of the two remaining pairs of parts 
must be symmetrical with respect to the other pairs of parts. For 
example, equation a remains unchanged when we interchange b 
and c, else it would be wrong. 

99. Theorem of sines. In a spherical triangle the sines of 
the sides are proportional to the sines of the opposite angles. 



FUNDAMENTAL PRINCIPLES. Q9 

Proof. Let O-ABC be the trihedral angle of the spherical tri- 
triangle, and let A be any point on ^ 

the edge A. ^ 

Through A pass a plane perpen- ^^ 

dicular to the edge OB, intersecting s^ I 

the faces A OB and BOCin the lines 0< <i L 

AB and BP. \\ / 

Through A pass another plane \V,---'' P 

perpendicular to 00, intersecting the \. 

faces AOOmd CO B in the lines AC and OP. 

AP will then be the line of intersection of these planes. 

Because the planes ABP and ACP are perpendicular to the 
lines OB and OC respectively, they are each perpendicular to the 
plane BOC of these lines (Geom.) Therefore their line of inter- 
section, AP, is also perpendicular to this plane, and the triangles 
APB and APC are right-angled at P. Hence 

AP = AB sin ABP = AB sin B. 
Also, because AB is perpendicular to OB, 

AB = OA sin BOA = OA sin e. 
Therefore AP = (9.A sin c sin i?. 

We find in the same way 

AP = OA sin b sin C; 
whence 

sin c sin B = sin J sin (7, 
or 

sin c sin J 



sin C ~ sin i?' 
We may show in the same way, by permuting the parts, 
sin a sin b sin c 



(2) 



sin^l — sin B ~~ sin C 
The common value of these three ratios is called the modulus 
of the spherical triangle. 

100. The theorem of sines may also be obtained directly from 
the fundamental equations as follows : 

From the first fundamental equation (1) we obtain 
cos a — cos b cos c 



cos A — 



sin b sin c 



120 SPHERICAL TRIGONOMETRY. 

Squaring, 

a . cos 9 a — 2 cos a cos J cos c + cos 9 5 cos 9 c 

COS 9 .A = . . -, . a ! 

sin 6 sin c 

Hence sin 9 A = l- cos 3 .A = 

(1 — cos 9 b) (1 — cos 9 c) — cos 8 fl -j- 2 cos a cos J cos c— cos 9 5 cos 9 c 

sin 2 J sin 3 c 
_ 1 — cos 9 a — cos 9 b — cos 3 c -\-2 cos a cos 5 cos <? 
sin 3 b sin 2 <? 
Dividing by sin 9 a, 

sin 3 ^1 1 — cos 9 a — cos 9 b — cos 9 # + 2 cos a cos 5 cos <? 

sin 3 # ~~ sin 3 # sin 3 5 sin 3 c 

The second member of this equation is symmetrical with 
respect to a> b, and e, and so remains unchanged when these quan- 
tities are permuted among themselves. But if we derive the 

_ sin 3 B _ sin 3 (7 . _ . 

values oi . a , and -r-5 — irom the last two fundamental equa- 
sm sm g ^ 

tions, the results will be simple permutations of the last equation, 

and will therefore give the same values of . a -, and -=-= — that 
sin sm c 

we have found for -r-= — . 
sin a 

Hence . 2 . . . n . , „ 

sm 9 J. sm 9 i? sm 9 tf 



sm 3 # ~ sin 3 b sin 9 <? ' 
Extracting the square roots, the general results will have double 
(±) algebraic signs; but as the angles are all supposed to be less 
than 180°, the positive signs are to be taken. Hence 

sin A _ sin B sin C 

sin a ~~ sin b ~~ sin ' 
the reciprocal of the relations (2). 

101. Polar triangles. Def. When two triangles are so related 
that the vertices of the one are the poles of the sides of the other, 
the one is said to be the polar triangle of the other. 

It is shown in geometry that the relation of a triangle to its 
polar triangle is reciprocal ; that is, if Xand I^are two triangles, 
and Yis the polar triangle of X, then X is the polar triangle of 
Y. This reciprocity arises from the theorem : 




FUNDAMENTAL PRINCIPLES. 121 

If A, B, and C be the three poles of the sides QP, PP, and 
PQ of a triangle PQP, then P, Q, and P will be the poles of 
the sides BO, CA, and AB. 

This theorem is readily proved by the geometry of the sphere. 

Since every great circle has two poles, one at each end of a 
diameter, it follows that the three sides of a triangle have six 
poles in all. We may form a polar triangle to ABC hj taking 
either of the poles of AB, either of the poles of BC, and either of 
the poles of CA, and joining them by R 

arcs of great circles. Hence there are y" x 

eight possible polar triangles to every 
given triangle. To avoid doubt which 
triangle is to be chosen, we take for / 
each vertex of the polar triangle that 
pole of each side of the given triangle /' 
which is on the side toward the triangle. p ~~~"~" 

For example, if ABC is the given triangle, we take that pole 
of AB which is on the side toward C, and so with the other sides. 

Exercises. 

1. What must be the sides and angles of a triangle that it may 
coincide with its polar triangle ? 

2. Show that if each side of a triangle is greater than 90° the 
polar triangle will fall wholly inside of it, and if each side is less 
than 90° it will be wholly within its polar triangle. 

3. If two sides exceed 90° and the third side is less than 90°, 
what will be the character of the polar triangle, and how will it be 
situated relatively to the given one ? 

102. Use of the polar triangle. It is shown in geometry that 
each side of the polar triangle is the supplement of the opposite 
angle of the other, and vice versa. This principle is applied to find 
new relations between the parts of a triangle in the following way : 

1. We imagine ourselves to construct the polar of the given 
triangle. 

2. We write any or all the equations between the parts of the 
polar triangle. 



122 SPHERICAL TRIGONOMETRY. 

3. "We substitute in these equations the supplementary parts 
of the given triangle, and thus obtain equations between these 
parts. 

Let us put #', b\ c', 

A', B', C, 
the sides and opposite angles of the polar triangle. Since the 
general equations (1) are true for every triangle, they are true of 
this polar triangle. Hence 

cos a/ = cos V cos c' -f- sin V sin c' cos A'. 
But the polar triangle is so related to the original triangle that 





a' = 180° - A, 


A! — 180° -a; 




y = 180° - B, 


B' = 180° - b ; 




c > = 180° - C, 


C = 180° - c. 


Therefore 








cos a' = — cos A, 


cos A f = — cos # ; 




cos V = — cos B, 


cos jB' = — cos 5 ; 




cos c' == — cos 0, 


cos O f = — cos c ; 


and 


sin a' = sin A, 


sin ^1' = sin a ; 




etc. 


etc. 



Making these substitutions in the equations (1), we find 
cos A = — cos ^ cos (7 -|- sin ^ sin C cos & ; ] 
cos i? = — cos C cos JL + sin {7 sin ^1 cos J ; > (3) 

cos (7 = — cos A cos B -\- sin ^1 sin JB cos <?. J 
This process may be generalized thus : 

From every relation between the parts of a spherical triangle 
we may derive another relation by changing the cosine of each 
part into the negative of the cosine of the opposite part, and the 
sine of each part into the sine of the opposite part. 

But this relation will not always be different from the original 
one. If we apply the process to the equations (2), for instance, 
the same relations will be reproduced, each term being changed to 
its reciprocal. 

It is also to be remarked that the use of the polar triangle is 
not absolutely necessary to deduce the new relations (3), because 
they can all be obtained from the fundamental equations (1) by 
eliminating first b and c, then c and a, then a and b. But the use 



FUNDAMENTAL PRINCIPLES. 123 

of the polar triangle gives a much shorter and more elegant mode 
of deducing them. 

103. In the solution of spherical triangles we require for- 
mulas which shall express any three parts in terms of the remaining 
three ; the latter being supposed known, the former unknown. 

A set of such equations may be derived from the fundamental 
equations (1) by substituting in any one the value of the cosine of 
a side obtained from another. Let us substitute in the third the 
value of cos a from the first. We shall have 
cos c = cos 2 b cos c -j- sin b cos b sin c cos A -{-sin a sin b cos O. 
Transposing cos c, noting that 1 — cos 2 b = sin 2 b, and dividing by 
,sin b, we have 

= — sin b cos c -f- cos b sin c cos A ~\- sin a cos O, 
which gives 

sin a cos C = sin b cos c — cos b sin c cos A. (4) 

The theorem of sines (2) also gives 

sin a sin C = sin c sin A. (5) 

Comparing these two last equations with the first equation (1), 
we see that they form a set in which the second members contain 
only the parts b, c, and A, namely, two sides and the included 
angle ; while the first members contain the third side, a, and one 
of the angles adjacent to it, namely, G. 

Hence any two of these three equations may be used to find 
the side a and the angle O when b, c, and A are given. Since 
there are three equations where only two are necessary, there must 
be a relation between them, which we find as follows : 
The first members are 

cos a ; 
sin a cos O; 
sin a sin O. 
The sum of the squares of these quantities is 

cos 2 a -\- sin 2 a(cos 2 O -f- sin 2 C) = cos 2 a -\- sin 8 a = 1. 
The sum of the squares of the first members being identically 
equal to unity, the same should be true of the sum of the squares 
of the second members, which is 



124 



SPHERICAL TRIGONOMETRY. 



(cos b cos c -\- sin b sin c cos Af 
-|- (sin b cos — cos b sin c cos A) 2 
-[- sin 3 c sin 2 A. 
Forming these squares, we see that the product of the terms 
in the squares of the first two quantities cancel each other, leaving 
as the sum of the squares 

cos 2 b cos 2 c -f- sin 2 b sin 2 c cos 2 A 
-f- sin 2 b cos 2 + cos 2 b sin 2 c cos 2 A 
-f- sin 2 c sin 2 ^L, 
a sum which we readily find to reduce to unity. 

104. By permuting the sides b and c and their opposite 
angles, B and O, we obtain a similar set of equations for a and the 
adjacent angle B. Repeating the equations already found, we 
have the following set of five equations from which we may deter- 
mine #, B, and (7, when b, c, and A are given : 

sin a sin B = sin b sin A ; 
sin a cos B = cos 5 sin c — sin 5 cos c cos -4 ; 
sin a sin (7 = sin sin ^L ; > (6) 

sin # cos C = sin 5 cos c — cos 5 sin c cos J. ; 
cos # = cos J cos c -\- sin J sin c cos .A. 

We may write a set similar to this for each of the other sides 
of the triangle, namely : 

sin b sin C — sin c sin B ; 
sin b cos C — cos sin a — sin cos a cos i? ; 
sin 5 sin JL = sin a sin i? ; I. (7) 

sin 5 cos A = sin cos a — cos sin a cos B ; 
cos 5 = cos c cos a -f- sm c sm ^ cos B- 
sin c sin ^1 = sin a sin (7; 
sin g cos -4. = cos a sin 5 — sin a cos 5 cos C\ 
sin <? sin i? = sin b sin (7; (8) 

sin g cos i? = sin a cos J — cos a sin 5 cos C; 
cos <? = cos a cos 5 -)- sin a sin 5 cos (7. 
Then we obtain a similar set for the angles through the inter- 
vention of the polar triangle. Applying the set (6) to the polar 
triangle, it gives 



FUNDAMENTAL PRINCIPLES. 



125 



sin A sin b = sin i? sin # ; 
sin ^L cos b = cos i> sin (7 -f- sin i? cos (7 cos # 
sin JL sin c = sin £7 sin a ; I (9) 

sin A cos 6' = sin B cos (7 + cos B sin 6" cos # ; 
cos A = — cos B cos tf-f~ sin B sin (7 cos a. 
Applying these formulae to each of the angles in succession, we 
have a set of equations by which, when two angles and the in- 
cluded side are given, the remaining parts may be found. The 
following are the remaining formulae obtained by permutation : 
sin B sin c = sin O sin b ; 
sin B cos c = cos C sin A -\- sin O cos A cos J 
sin B sin & = sin .A sin 5 ; I (10) 

sin B cos a = sin 67 cos J. -f- cos 6" sin J. cos b ; 
cos B = — cos (7 cos JL -f- sin C sin ^1 cos J. 
sin C sin # = sin ^ sin c ; 

sin C cos a = cos A sin .Z? -f- sin .A cos i? cos c ; 
sin (7 sin b = sin .Z? sin c ; I (11) 

sin C cos J = sin A cos ^ -(- cos A sin J5 cos c ; 
cos O = — cos ^1 cos jB -|- sin ^1 sin i? cos c. 

105. If in the three sets of equations (6) to (8) we divide the 
second equation by the first, and the fourth by the third, and clear 
of denominators, we shall have the following additional equations : 

cot B sin A = cot b sin c — cos c cos A ; 
cot C sin A = cot c sin 5 — cos b cos A ; 
cot (7 sin B = cot c sin ^ — cos a cos B ; 
cot u4 sin _Z? = cot & sin c — cos c cos B ; 
cot J. sin O = cot a sin b — cos J cos O ; 
cot .Z? sin C = cot 5 sin a — cos # cos (7. 

Treating the equations (9) to (11) in the same way, we shall get 
a similar set of equations ; but on examination they will be found 
to differ from the equations (12) only in the arrangement of their 
terms, so that they need not be written. 

106. Although the foregoing equations need to be trans- 
formed for most of their uses, they may in many cases be applied 



(12) 



126 SPHERICAL TRIGONOMETRY. 

directly to the solution of spherical triangles. This is especially the 
case when two sides and the included angle, or two angles and the 
included side, are given and one or more of the remaining parts 
required. If, for instance, A, B, and c are given, we may from are 
the first two of equations (11) obtain the values of O and a by 
Prob. I. Chap. V. Then from the next two we may obtain c and 
(7, and from the last cos C, and thus a third value of C. If the work 
is correct, these three values of O will all agree. 

Example. Two of the face-angles of a trihedral angle are a — 
132° 46'.7 and b = 59° 50'.1, and the included edge-angle is G = 
56° 28'.4. Find the remaining parts. 

The computation of A, B, and c may be effected by the equa- 
tions (8), as follows : 

(1) cos b, 9.701 13 (6) sin b, 9.936 81 

(2) sin a, 9.865 69 (7) cos G, 9.742 19 

(3) sin C, 9.920 97 (8) cos b, 9.701 13 

(4) sin b, 9.936 81 (9) sin a, 9.865 69 

(5) cos a, — 9.831 98 (11) sin b cos <7, 9.679 00 
(12) cos a sin b, - 9.768 79 (10) cos a, - 9.831 98 

(13) -sinacos&costf, - 9.309 01 (20) sin a cos b, 9.566 82 

(14) Diff., 0.459 78 ( 21 ) -cosa sin&cos<7, + 9.51 98 

(15) Add. log., 0.589 12 (22) Diff., 0.055 84 

(2) + (3) sin c sin A, 9.786 Q6 (23) Add. log., 0.329 84 

(13)+(15) sine cos A, - 9.898 13 (26) sin B, 9.857 80 

(18) cos A, 9.898 14 (3)+(4) sin c sin B, 9.857 78 

(16) tan A, - <^888~53 (21)+(23) sin c cosB, 9. 840 82 

(17) A, 142° 16 / .4 (24) tan B, 0.016 96 

(19) sin c, 9.999 99 (25) B, 46° 7'.1 

(27) sin c, 9.999 98 

(31) Subtr. log., 8.4317 (30) Diff., 0.01158 

(28) cob a cob b, -9.533 11 ^— 

(29) sin a sin b cos G, 9.544 69 QQO , 

C, 0*7 .Zo .O 

Note.— We may omit the "log" from the designation of the logarithms of 
the trigonometric functions whenever no uncertainty will thus arise. 



FUNDAMENTAL PRINCIPLES. 



127 



In this computation the numbers in parentheses show the order 
in which the lines may be written. Lines (15), (23), and (31) 
are the addition or subtraction logarithms, from the tables for 
finding the logarithm of the sum or difference of two numbers 
which are given by their logarithms. The student can equally 
well find the numbers, add them together, and take the logarithm 
of their sum. 

The agreement of the two values of sin G with each other and 
with cos G shows the correctness of the calculation. 

107. The following transformation, similar to that of Prob. 
IY. Chap. V., will often render the work convenient. In the 
second and last of equations (11) let us put 

Jc sin K = sin A cos c ; ) 
JfccosJT= -cosJL \ W 

By substitution these equations will then become 

sin {7 cos a = — Jc cos iT sin B -f- Jc sin iTcos B 
= &sin(^- B); 
cos G = Jc cos iT cos B -f- h sin iTsin B 
= Jcco$(K- B). 

To apply these equations we compute Jc and K from (a), and 
then sin G cos a and cos G from (5). "We complete the work by 
computing sin G sin a from the first equation of (11). 

"We may also transform the fourth equation by computing h 
and H from the equations 

h sin H = sin i? cos c ; 
h qosH = — cos .Z?. 
We shall then have 

sin (7 cos 5 = h sin (5" — A) ; 
cos 6" = h cos (J5T — -4). 
To transform the equations (8) on the same plan, we may com- 
pute k, K, h and .ZTfrom 

Jc sin K = sin a cos G; ' 
& cos iT = cos a ; 
7^ sin jS" = sin 5 cos G 
h cos iT" = cos b. 



© 



(o) 



128 SPHERICAL TRIGONOMETRY. 

We then have, in the same way as before, 

sin g cos A = Tc sin (b — K) ; (d) 

sin g cos B = A sin (a — H) ; 

cos G = k cos (b — K) = A cos (a — iJT). 
"We compute the same example as before by these formulae, as 
follows : 

sin a, 9.865 69 
sin C 9 9.920 97 
sin b, 9.936 81 

b-K - 89° 19'.9 
a - H, 89° 14'.3 

sin (b-K), - 9.999 97 
log k, 9.898 16 

cob(& — J5T), 8.066 9 



sin #, 


9.865 69 


cos C, 


9.742 19 


sin 5, 


9.936 81 


& sin K, 


9.607 88 


Tc cos iT, - 


- 9.831 98 


cos K, ■ 


- 9.933 82 


tan i£5 • 


- 9.775 90 


K, 


149° lO'.O 



A sin H, 
A cos H, 


59° 50'.1 
9.679 00 
9.701 13 

9.860 27 


sin g sin A, 
sin g cos A, ■ 
cos JT, ■ 

tan A y 
A, 

sin g, 

sin (a — H), 
log A, 

sin g sin B, 
sin <? cos B, 

tan j5, 


9.786 66 

- 9.898 13 

- 9.898 14 


cos H, 


- 9.888 53 


tan H, 


9.977 87 

43° 32'.4 

132° 46'.7 

7.965 1 
7.964 5 

89° 28'.3 

46°r.i 


142° 16'.4 
9.999 99 


a, 
cos c, 

COS G, 


9.999 96 
9.840 86 
8.123 6 




9.857 78 
9.840 82 




0.016 96 



Exercises. 

1. Transform the equations (6), (7), (9), and (10) in the same 
way that we have transformed (8) and (11). 

2. From the values of A, B, and c, which we have obtained 
in the last example, find those of a, b, and C with which we started. 

3. If m be the arc joining the vertex A to the opposite side, 

prove 

cos b -\- cos g = 2 cos \a cos m. 



CHAPTER II. 

RIGHT AND QUADRANTAL IRIANGLES. 



Fundamental Definitions and Theorems. 

108. Def. A right spherical triangle is one which has a 
right angle. 

Def A quadrantal spherical triangle is one which has a side 
equal to a quadrant. 

Def. A trirectangular triangle is one which has three right 
angles. 

Def. A birectangular triangle is one which has two right 
angles. 

Def. A biquadrantal triangle is one which has two sides equal 
to a quadrant. 

Theorem I. Every birectcmgular triangle is also biquad- 
rantal. 

Proof. Let ABO be a spherical triangle in 
which angle B = angle O = 90°. Then : 

Because angle B is a right angle, the pole of 
the great circle BO is on the great circle BA. 

Because angle is a right angle, this pole is 
on the great circle OA. (Geom.) 

Therefore the pole of BO is on both BA and OA, and there- 
fore at their point of intersection A. 

Because A is the pole of BO, AB and A O are quadrants. 

Q.E.D. 

Theorem II. Conversely, Every biquadrantal triangle is also 
birectangular. 

Proof. Because every point of the polar circle of the point 
A is a quadrant distant from A, and because AB and A O are 
quadrants, this polar circle must pass through both B and O. 




130 SPHERICAL TRIGONOMETRY. 

But only one great circle can pass through these points. 

Therefore BO is the polar circle of A, and A the pole of BO. 

Therefore the great circles AB and AG intersect BO at right 
angles. Q.E.D. 

Oor. Every trirectangulor triangle has three quadrants for 
its sides ; and, 

Conversely, Every triangle having three quadrants for its 
sides is trirectangular. 

TheopwEM III. In a birectangular triangle the oblique angle 
is equal to its opposite side. 

Proof. Because the plane of the great circle BO intersects 
the planes of AB and of A at right angles, the arc BO measures 
the dihedral angle between the planes AB and BO. 

But the angle A is equal to this same dihedral angle. 

Therefore BO = angle A. 

Theorem IV. The polar triangle of a right triangle is a 
quadrantal triangle. 

This follows at once from the fact that the angles of the one 
triangle are the supplements of the sides of the other. 

Exercise. 

Let the student translate the preceding definitions and theorems 
into those relating to the face- and edge-angles of a trihedral angle, 
and, which is the same thing, into those relating to the angles be- 
tween three lines emanating from a point and the angles between 
their planes. 

109. Formulae for right triangles. Since in a right triangle 
one of the parts, the right angle, is known in advance, if two other 
parts be given the remaining three parts may A 

be found. c. 

An equation must therefore exist by 

which, when any two parts are given, any g. 

one of the three remaining parts may be 

found ; hence between every combination of three parts out of the 

five there must be an equation. The number of combinations of 3 

5 . 4 
in 5 being j^-= = 10, there must be ten such equations. 




RIGHT AND QUADRANTAL TRIANGLES. 



131 



To find these equations let C be the right angle, and therefore c 
the hypothenuse. We seek for those equations in the sets (6) to 
(12) of the last chapter, in which the angle C enters, and in which 
the equation contains only three different parts. We then suppose 

sin O = 1 ; 



cos C — ; 



cot (7=0. 
The set from which the required equations are taken, the num- 
ber in the set, and the result are shown as follows : 



From (6) 3 , sin a = sin c sin A ; 

(7)„ sin b = sin c sin B ; 

(6) 4 , cos A = tan b cot c ; 

(7) 2 , cos B = tan a cot c ; 

(8) 6 , cos c = cos a cos b ; 

(9) 3 , cos B = cosb sin ^1 ; 

(9) B , cos A = cos asinB ; 
(11) 5 , cos c = cot A cot ,5 ; 
(12) 5 , cot A = cot & sin & ; 
(12) 6 , cot B = cotb sin #. 



CD} 

(3)) 
(*)) 

(5) 
(6)| 

(W 

(8) 
(9)1 
(10) J 



These ten equations will be found to include all combinations 

of three out of the five parts a, b, c, A, B. From each of them 

we may determine any one part in terms t f the other two ; for 

example, the first equation gives not only 

sin a = sin c sin A, 
but 



and 



sin c 



sin A = 



sin a 
sin A 9 

sin a 



sin c 



Properly speaking, only six of these equations are really dis- 
tinct, as the other four can be derived from them by a mere inter- 
change of letters between corresponding parts. For instance, since 
the same relation must hold between each oblique angle and its 
opposite side, the second equation may be derived from the first. 

The equations which are thus related are connected by braces 
in the formulse above. 




132 SPHERICAL TRIGONOMETRY. 

110. Napier's rules. The six preceding formulae, which may 
be found difficult to remember, have been included by Napier in 
two precepts of remarkable simplicity, 
and easily remembered. 

Let us take for the five parts the 
sides a and b as before, and, instead of 
the other three parts, the complements 
of the oblique angles and of the hy- 
pothenuse. The fact that the comple- 
ments are understood is indicated by accenting the letters in the 
diagram. We suppose 

B' = 90° - B; c' = 90° - c; A' = 90° - a. 
Omitting the right angle, the five parts a, b, A!, c r , B' form a con- 
tinuous series, B' being followed in regular order by a. Now if 
we select any three of these parts, one of two cases must occur. 
Either— 

(1) The three parts all adjoin each other, as B f , a,b; a, h, A', 
etc., or 

(2) Two of the parts adjoin each other and the third is separated 
from each of them by the remaining intervening parts. 

The middle part of the three in the first case, or the separated 
part in the second, is called the middle part 

In the first case the extreme parts of the three are called 
adjacent parts. 

In the second case the adjoining parts are called opposite parts. 

111. Napier's rules are : 

I. The sine of the middle part equals the product of the tan- 
gents of the adjacent parts. 

II. The sine of the middle part equals the product of the 
cosines of the opposite parts. 

The concurrence of the vowel a in tangent and adjacent, and of 
the vowel o in cosine and opposite, will help in remembering the 
relations. 

Examples. 1. Let the parts be the hypothenuse and the two 
adjacent angles, or c, A and B. 

The middle part is c', and A' and B' are adjacent parts. 



RIGHT AND QUADRANTAL TRIANGLES. 133 

By the rule, 

sin (90° - O) = tan (90° - A) tan (90° - B), 
or cos c = cot A cot B f 

agreeing with the formula (8). 

2. Let the parts be a, A and C. The middle part is then a y 
and A' and C are opposite parts. Therefore 

sin a = cos (90° - A) cos (90° - C) 
= sin .A sin c, 
agreeing with the formula (1). 

3. Let the three parts be the two sides containing the right 
angle and one of the oblique angles, say a, b, and A. Then b is 
the middle part, and the other two adjacent parts. Therefore 

sin b = tan a tan (90° — A), 
or sin b cot a = cot A, 

agreeing with the ninth formula. 

Exercises. 

1. Given A = 62° 29 / .3, b = 25° 58'.8 ; find a. 

2. Given B = 35° 29'.6, a = 75° 5 ; .3 ; find A and b. 

3. Given a = 43° 40'.5, c = 98° 29 7 .1 ; find A, B, and J. 

4. Given a = 148° 28\2, A = 101° 3'.9 ; find 5 and <?. 

5. Given A = 50° 0'.8, B = 79° 57 r .3 ; find a, 5, and c. 

112. Relations between four parts. Although the preceding 
formulae enable us, when two parts are given, to find the remain- 
ing three parts, each part has to be found independently by 
different equations. If all three parts are required, we may deter- 
mine two of them by a single connected set of operations. For 
this purpose we select the appropriate equations from the sets (6) 
to (11) of the preceding chapter, choosing only those in which the 
angle Centers the second member. 

113. Case I. Given an angle and the adjacent side. 
When C = 90°, the last three equations (9) of § 104 are 

sin A sin c = sin a ; ] 

sin A cos o = cos a cos B ; [ (11) 

cos A = cos a sin B. J 



134 SPHERICAL TRIGONOMETRY. 

From the first two equations we obtain sin A and c. Since we 
thus have separate values of sin A and cos A, the agreement of 
the two values of A serves as a check upon the accuracy of the 
computation. 

If ft is also required, the first two equations (9) of § 104 give 

sin A sin ft = sin a sin B ; ) n ~ 

sin A cos ft = cos B. ) 

From which we obtain ft and another value of sin A, 

By simply reversing equations (11) we obtain a and B when 
A and c are given. 

Example. Given a = 75° 5'.3, ^ = 35° 29'.6, to find the 
remaining three parts. 

sin a, 9.985 12 

sin ^, 9.763 88 

cos a, 9.410 49 

cos.#, 9.910 73 

sin A sm c = sin #, 9.985 12 
sin A cos c = cos a cos i?, 9.321 22 

tan c, 0.663 90 c = 77° 46'.0 

sin^L, 9.99510 
cos A = cos a sin _£, 9.174 37 A = 81° 24'.4 



sin A sin J, 9.749 00 
sin A cos ft, 9.910 73 

tan 5, 9.838 27 b = 34°34'.2 
sin A, 9.995 10 

EXEECISES. 

1. Given a = 34° 34'.2, B = 81° 24'.4 ; find remaining parts. 

2. Given A = 45° 45'.4, c = 61° 49'.3 ; find a and ^. 

3. Given a = 120° 29'.6, B = 22° 59'.8 ; find c and JL. 

4. Given A = 98° 0'.4, ft = 52° 7'.8 ; find a and .#. 

5. Given B = 133° 33'.7, a = 7° 29'.3 ; find J. and 5. 
114. Case II. Given the two sides, a and ft. 
Putting C = 90°, the equations (8) of § 104 become 



RIGHT AND QUADRANTAL TRIANGLES. 



135 



sin c sin A = sin a ; 
sin c cos A = cos a sin b ; 
cos c = cos # cos 5 ; 



(13) 



sin c sin B = sin 5 ; 

sin c cos i? = sin & cos J. 
The values of c and J. are determined from the first three 
equations ; those of c and B from the last three. The agreement 
of the two values of sin c with the one value of cosine c affords a 
check upon the accuracy of the work. 

Exercises. 

1. Given a = 39° 6'.8, b = 82° 39'.6 ; find A, B, c. 

2. Given a = 103° 40'.2, 5 = 62° 29'.3 ; find A, B, c. 

3. Given a = 172° 1'.5, 5 = 158° 58'.8 ; find A, B, e. 

115. Case III. Given the hypothenuse and one angle. 

The first three equations of Case I. and the first three of 

Case II. give 

cos a sin B = cos A ; 

cos a cos B = sin A cos c ; 

sin a = sin A sin c ; [ (14) 

cos a sin 5 = cos ^4 sin c ; 

cos a cos 5 == cos c. 

From which a, b, and i? may be determined. 

Exeecises. 

In a triangle, right-angled at C, prove the relations : 

1. sin A sin 2b = sin c sin 2^. 

2. sin 2 A sin c = sin 2a sin J?. 

3. sin 2a sin 25 = 4 cos J. cos B sin 2 c. 

4. sin 8 \c = sin 2 -|& cos 2 %b -f- sin 2 £5 cos 2 \a. 

5. sin (c — b) — tan 2 -§-J. sin (5 + c). 

6. sin a cos b = tan -JJ. sin (b -\- c). 

7. In a right triangle of which the oblique angles are 

A = 69° 23'.7, B = 60° 7'.6, 
find the length of the perpendicular from the right angle upon the 
base, and the angles which it forms with the sides. 



136 SPHERICAL TRIGONOMETRY. 

8. In a right triangle is given 

c = 75° 25', a = 52° 16'; 
find the lengths of the segments into which a is divided by the 
bisector of A. 

Exercises in Geometric Application. 

9. From a point P above a plane an oblique line PO is 
drawn, meeting the plane in and making the angle A with the 
plane. Let Q be the projection of P upon the plane, so that OQ 
is the projection of OP. Through O a line OMis drawn, making 
an angle QOM — B with the projection OQ. It is required to 
express the angle POM in terms of A and B. 

Ans. Cos POM = cos A cos B. 

10. In the preceding case, if a perpendicular PS be dropped 
from P upon OM, express the length OS in terms of the angle A 
and B and the length OP. Ans. OS = OP cos A cos B. 

11. Two planes intersect at right angles along a line /. From 
any point P oi I one line is drawn in each plane, making the 
respective angles A and B with I. Express the angle C between 
these lines. Ans. Cos C = cos A cos B. 

12. Two planes intersecting at right angles along a line I are 
intersected by a third plane, making with them the respective 
angles P and Q. Express the angles which the three lines of in- 
tersection make with each other. 

Ans. If we put PI for the angle between /and that edge 
along which the dihedral angle P is formed, etc., we have 

cos Q 



cos PI 
cos QI = 



sinP' 
cosP 



sin Q> 
cos PQ = cot P cot Q. 

116. Isosceles triangles. An isosceles spherical triangle may 
be divided into two symmetrical right triangles by a perpendicular 
from its vertex upon its base. If we put 
e, each of the equal sides ; 
O, each of the equal angles at the base ; 



EIGHT AND QUADRANTAL TRIANGLES L37 

b, the base, or third side ; 
B, the angle at the vertex ; 

1\ the middle point of b ; 

<p, the length of the perpendicular BP from B upon b, — 
we shall then have two right triangles in each of which the 
oblique angles are O and %B, the hypothenuse is c, and the sides 
containing the right angle are p and %b. The equations of § 109 
will then give 

sin %b — sin c sin \B ; 

sin jp = sin c sin (7 ; 

etc. etc. 

Exercises. 

1. The equal sides of an isosceles triangle are each 45°, and the 
angle which they contain is 95°. Find the base and the angles at 
the base. 

2. If the base of an isosceles triangle is 95°, and the angles at 
the base each 45°, find the remaining parts. 

117. Quadrantal triangles. Since the polar of a right tri- 
angle is a quadrantal triangle, the formulae for quadrantal triangles 
may be obtained by applying the formulae of § 109 to the polar 
triangle. The side c will then be a quadrant, and the relations 
among the other parts will be 

sin A = sin C sin a ; 

cos a = — tan B cot G\ 

cos C = — cos A cos B 
cos b = cos B sin a ; 

cos G = — cot a cot b ; 

sin B = tan A cot a. 

If we take, as the five parts of the triangle, 

A, B, 90° - a, 90° -l,G— 90°, (a) 

and omit the hypothenuse c, the above formulae will be expressed 
by a set of rules identical in expression with those of Napier. For 
example, let us consider the parts a, b, C. Here C will be a 



(15) 



138 SPHERICAL TRIGONOMETRY. 

middle part, and a and h adjacent parts. Applying Napier's rules 
to this case, with the parts (a) we have 

sin (0- 90°) = tan (90° — a) tan (90° - V) ; 
which gives — cos C = cot a cot 5, 

an equation identical with the fifth of the above list. 

Exercises. 

1. Let the student deduce the six equations (15) by applying 
Napier's rules to the parts (a). 

2. Through the same point there pass two lines intersecting at 
right angles, and a plane P making the angle a with one of the 
lines, and the angle /? with the other. Express the angle which 
the plane P forms with the plane of the lines. 

Ans. Sin A = Vsin 2 a -f- sin 2 p. 

3. The sides of an obelisk have a slope of 8° from the perpen- 
dicular. What is the face-angle at the base of the obelisk, the 
slope of the edges, and the dihedral angle between two adjacent 
lateral faces? Ans. Face-angle at base, 82° 4/.6 ; slope, 11° 14'.5 ; 
dihedral angle, 91° 6'.6. 

In this problem, to reduce to a spherical triangle, consider the centre of the 
sphere to be at a corner of the obelisk. The slope of the edge will not be 
represented by either of the six parts of the triangle, but by the complement of 
the perpendicular from the vertex upon the base. 

4. A mason cuts a stone with a rectangular base and four lateral 
edges, each making an angle of 60° with the base at its corners. 
What is the inclination of each lateral face to the base, and the 
dihedral angle between the faces, supposing such inclinations and 
dihedral angles all equal ? Ans. 67° 47'.5 and 98° 12 / .8. 

5. In another stone the base is rectangular; one lateral face 
makes an angle of 68° 29' with the base, and the lateral edge 
bounding this face makes an angle of 52° 15' with the base. What 
angles does the adjacent lateral face make with the first face and 
with the base ? 

6. When the angular distance of the sun from the south point 
of the horizon is 75°, and from the west point 60°, what is its alti- 
tude above the horizon ? 



CHAPTER III. 

TRANSFORMATION OF THE FORMULA OB SPHERICAL 
TRIGONOMETRY. 



118. Although the formulae already given suffice for the solu- 
tion of every spherical triangle, there are many transformations 
which will facilitate the applications of spherical trigonometry, 
and render the solutions of triangles more accurate and convenient. 
Let us first take the fundamental equation (1) of Chapter I., 
cos a = cos b cos c -f- sin b sin c cos A. 
We may express this in the form 

cos a = cos b cos c -f- sin b sin c — sin b sin c (1 — cos A) 
= cos (b — c) — 2 sin b sin c sin 8 \A. 
Moreover, because 

— 2 sin b sin c == cos (b + c) — cos (b — <?), (§ 43) 

we have, by substituting, 
cos a = cos (5 — c) (1 — sin 3 %A) -f- cos (3 -f- c) sin 2 -JJL 
= cos (b — c) cos 2 \A -\- cos (5 -f- c) sin 2 JJ.. 
This last equation may also be derived by the following elegant 
process. The original fundamental equation may be written 
cos a = cos b cos c (cos 2 \A -\- sin 2 %A) 
-f- sin b sin o (cos 2 \A — sin 2 -|^1) 
(because cos 2 \A -f- sin 2 -J A = 1, and cos 2 \A — sin 2 J- J. = cos A). 
By conjoining the coefficients of cos 2 \A and of sin 2 %A, the 
equation (2) follows by the addition theorem. 
By a similar process, from the equation 

cos A — — cos B cos -f- sin B sin C cos #, 
we obtain 

cos A = — cos (B -j- (7) — 2 sin i? sin (7 sin 2 \a ; (3) 

cos .A = — cos (B -f- <7) cos 2 J& — cos (B — (7) sin 2 J&. (4) 
By a slight modification of the process employed in forming 
the equations (1) and (3) we may find 



i« 



| (2) 



140 



SPHERICAL TRIGONOMETRY. 



cos a = cos (b -f- c) -f- 2 sin 5 sin c cos 2 -^4 ; (5) 

cos A = — cos (B — tf) -|- 2 sin i? sin (7 cos 8 ■£# ; (6) 

which equations the student may prove as an exercise. 

119. Expressions when three sides or three cmgles wre given* 

From the last equation (1) we find 

. „ „ . cos (b — c) — cos a 

sm 2 \A = o • / • ; 

2 sin b sm c ' 

by which any angle is expressed in terms of the three sides. 

By § 44, 13 we have 

cos (J> — c) — cos a = 2 sin %(a -f- c — 5) sin J (a -(- & — c). (a) 

If we put 5 for half the sum of the sides, namely, 

s = i(a + b + c), 
we have 

%(a -{- c — b) = s — b 

i(a -f- b — c) — s — c. 

Substituting these values in (a), the expression for sin a $A 

becomes 

sin (s — b) sin (s — c) 



:1 



Q>) 



sin 2 \ A 
then, by permutation, 



sin b sin c 

sin (5 — c) sin (s — a) fc 
sm 2 £i? = ; — — : ; 



sm c sm & 



sin 3 iO = 



sin (5 — a) sin (5 — b) 
sin # sin 5 



(7) 



To find similar expressions for the cosines we take equation 

(5), which gives 

cos a — cos (b -f- c) 



cos 8 \A 



2 sin b sin £ 



But 



cos a — cos (J -f" c) = 2 sin £(5 -f* c -f- a) sin i{b-\-c— a) 

[from (J)]. 
Therefore 



cos 3 \A = 
cos 3 J^ = 
cos 8 \G = 



= 2 sin s sin (5 — a) 

sin s sin (5 — a) 

sin b sin c 
sin s sin (5 — J) 

sin c sin a 
sin s sin (5 — c) 

sin a sin 5 



(8) 



TRANSFORMATION OF THE FORMULA. 

i 



141 



Since an angle near 90° cannot be accurately determinea by its 
sine, nor one near 0° by its cosine, neither of the formulae (7) or 
(S) can be advantageously used in all cases. But by taking the 
quotient of each equation (7) by the corresponding one of (8) we 
have 



tan 2 i A = 
tan' iB = 
tan 2 iC = 



(9) 



sin (s — b) sin (s — c) " 

sin s sin (s — a) 
sin (s — c) sin (s — a) 

sin s sin (s — b) 
sin (s — a) sin (s — b) 

sin s sin (s — c) 

120. Treating the equations (3) and (6) in the same way, and 

putting S=i(A + B+ C), 

we find the following expressions for the sides, in terms of the 

three angles : 

. , , cos/S'cos(xS r — A) 
sin 2 \a = — ; — ^r- — 77-^ ; 

* sm B sin O ' 

. , , _ cos S cos (S — B) 
sm 2 %b = — = — 77- — - A — - ; 

* sm C sm A ' 

cos 8 cos (# — (7) 
sin A sin J? 



sm' *c = 



(10) 



cos 2 £& = 
cos 2 ^» = 



cos (S-B) cos (£-(7) 
sin i? sin C 

cos(S-C)cos(S-A) 
sin (7 sin A 



cos 2 £c = 

tan 2 \a = — 
tan 2 P = - 
tan 2 £<? = — 



cos{S-A)cos(S-B) 



(11) 



sin A sin B 

cos # cos (# — A) 
cos(S-B)cos(S-Cy 

cos # cos (S — B) 
cos(S- G) cos (S — Ay 

COS ytfcOS (£ — 6') 

cos(£- J.) cos (£ -- .#)' 



(12) 



For the solution of a triangle in which all three sides or all 
three angles are given, the equations (9) and (12) are preferable. 



142 



SPHERICAL TRIGONOMETRY. 



For convenience in compntation the following slight modification 
may be made. Put • 

_ a/ em (s — a) sin (s — b) sin (s — c) . 

then from (9), 

tan 8 * A 



sin s 



V 

sin 2 (s — a) ' 

extracting the square root, and writing the remaining equations, 

P 



tan -J A = 
tan \B = 
tan \C 



sin {s — a) ' 

__£___. 

sin («§ — J) ' 



sin (s 
In the same way, if we put 



o) 



(13) 



P=y. 



cos # 



cos (tf - ^L) cos (£ - B) cos (# - C)> 



we find, from (12), 



tan \a = P cos (# — -<4) ; \ 
tan # = P cos (S-B)'A 
tan Jc = P cos (8 — C). ) 

Example of Computation. 
Given the three sides, 



(14) 



a = 76° 29'.3, 
find the angles. 

a= 76°29'.3 
5 = 93° 18'.6 
c = 122° 7'.7 



& = 93°18'.6, e— 122° r.7; 



5 = 145° 57'.8 
5 - a = 69° 28'.5 
5 - I — 52° 39 / .2 
2^ = 291°55 / .6 s — c = 23° 50'.1 



cosec 0.252 02 

sin 9.971 52 

9.900 36 

9.606 49 



9.865 20 

sin (s — a), 9.971 52 

tan \A, 9.893 68 

\A, 38° 3'.35 

A, 76° 6'. 7 



9.865 20 

sin (s — l\ 9.900 36 

tani^, 9.964 84 

\B, 42°41 / .0 

^, 85° 22'.0 



\ogp% 9.730 39 

log i?, 9.865 20 

sin (s—e), 9.606 49 

tan \G, 0.258 71 

\C, 61° 8'.3 

tf, 122° 16'.6 



TRANSFORMATION OF THE FORMULA. 143 

We may now check the results by computing the values of 
sin a sin b sin c , . _ . 

snTZ' shTi?' shT6 y ' ai Seeing whetlier the 7 a S ree - We nud 
sin a = 9.987 81 ; sin b = 9.999 27 ; sin c = 9.927 81 
sin A = 9.987 11 ; sin B = 9.998 58 ; sin (7 = 9.927 10 

.000 70 .000 69 .000 71 

Although the agreement is not perfect, the deviations are no 
greater than those due to the omission of decimals in the logarithms 
and angles. The angle B is that for which the check is most 
doubtful, because it is so near 90° that it may be changed V or 2' 
without changing the last figures in log. sin. by more than one or 
two units in the fifth place. In such a case certainty can be 
reached only by a duplicate computation. 

Exercises. 

1. Given a = 105° 6'.8, b = 93° 39'.9, c = 50° 20'.3 ; find the 
angles. 

2. Given A = 46° 59 , .3, B = 122°32 / .6, C = 139° r .3 ; find 
the sides. 

3. Given A = 78° 40'.7, B = 102° 29 r .5, O = 86° 49 / .4; find 
the sides. 

4. If the sides of a spherical triangle are each 120°, find the 
following numerical expressions for the sine, cosine, and tangent 
of each of the angles, and thence, by the aid of the tables, find 
the angles themselves. 

sines = V\ ; 

cosines = Vi ; 
tangents = V2. 

121. CagnolVs equation. Cagnoli's equation is 
sin a sin b -\- cos a cos b cos C = sin A sin B— cos A cos B cos c. (15) 

Proof. Multiplying the third equation (§ 97, 1) by cos C, we 
have 

cos c cos C = cos a cos b cos C + sin a sin 5 cos 8 (7 ) 

= cos a cos 5 cos £7 -f- sin <z sin b — sin # sin b sin 2 (7. j 
The equation of sines (§ 99, 2) gives 



144 SPHERICAL TRIGONOMETRY. 

sin a sin G = sin A sin c ; 

sin 5 sin G = sin i? sin c ; 

whence, by multiplying, 

sin a sin J sin 2 C = sin A sin i? sin 2 c 

= sin ^4 sin B — sin .4. sin B cos 9 <?. 

Substituting this value in (a), and interchanging the terms of the 

equation, 

sin a sin b -f- cos a cos J cos G 

= cos <? cos G + sin ^4 sin i? — sin JL sin B cos 2 e 
= sin A sin B -\- cos c (cos C— sin A sin j5 cos c). 
From the last equation (11), § 104, we have 

cos G — sin A sin i? cos c = — cos A cos i?. 
Making this substitution, we have the equation (15) as enunciated. 
Of course two other similar equations may be obtained by per- 
muting the symbols. 

122. Gauss's equations. We write the four equations, 

(a) Cagnoli's equation, 

(b) and (e) the fundamental equations (1) 3 and (3) 3 , and 
(d) the identical equation 1 = 1, as follows : 

(a) sin a sin b + cos a cos b cos G — sin A sin B — cos A cos J9 cos c ; 

(5) cos a cos 5 -f- sin a sin 5 cos (7 = cos c ; 

(<?) cos G = — cos .A cos .Z? -|- sin ^4 sin Z? cos <? ; 

(*> 1 = 1. 

Taking the sum of the four equations, and substituting 

cos (a — b) = cos a cosb -\- sin & sin 5, 

cos (A-\- B) = cos ^4 cos B — sin ^4 sin B, 

we have 

cos (a — b) + cos (7 cos (a — b) -f- cos (7+ 1 = 

— cos (J. -f- j5)— cos c cos (A -\- B) -\- cos c -\- 1 ; 
or 

(1 + cos G) [1 + cos (a - b)] = (1 .+ cos c) [1 - cos (A + B)] (e) 

If we form the sum — (a) — (b) -f- (o) -f (<#), and reduce in 
the same way, we find 

(1 + cos G) [1 - cos (a-b)] = (1 - cose) [1 - cos(A-B)] (f) 

The sum - (a) + (b) - (c) + (d) gives 



(16) 



TRANSFORMATION OF THE FORMULAE. 145 

(1 _ C os C) [1 + cos (a + b)] = (1 + cose) [1 + cos (.4 + B)] (g) 

The sum (a) — (b) — (c) + (d) gives 
(1 - cos G) [1 - cos (a + b)] = (1 - cos c) [1 + cos (A - B)] (h) 
In the equations (<?), (/), (#/), (A) we substitute the values of 
1 ± cos, namely, 

1-fcos 0= 2 cos 2 iC; 
l_cos <7=2sin 2 J<7; 
etc. etc., 
and dividing by 2, we obtain 

sin 2 \G sin 2 \{a + b) = sin 2 Jc cos 2 i(A - B) ; (A) 
sin 2 ££7 cos 2 \{a + b) = cos 2 Jo cos 2 J(^L + i?) ; (#) 
cos 2 i^sin 2 J(a -b) = sin 2 Jo sin 2 J(^. -B);(f) 
cos 2 Jtfcos 2 J(a — b) = cos 2 Jc sin 2 J(^L + B). (e) 
Extracting the square roots, we have 

sin \G sin \{a -f- b) = sin \c cos \{A — B) ; 
sin \G cos i(a -\- b) = cos J<? cos %(A -f- B) ; 
cos i sin J(a — b) = sin Jc sin i(A — B) ; 
cos \G cos 4(0 — b) = cos Jc sin J(^L + .Z?). 
In strictness, the second members of this equation may have 
the negative as well as the positive sign. But it is easy to show 
that when the sides and angles are all less than 180°, all the mem- 
bers of the equations are positive. Hence the positive sign is the 
only one necessary to be taken into account. 

These equations are applicable when any three consecutive 
parts of the triangle — two angles and the included side, or two 
sides and the included angle — are given. 

In the first case the three given parts are all in the right-hand 
members of the equations ; in the second case they are all on the 
left. 

These equations are written in the most convenient order for 
use in the first case ; in the second, the following is the order and 
arrangement : 



sin \g sin J( A — B) = cos \ G sin \{a — b) 
sin \c cos \(A — B) — sin J G sin J(# -\- b) 
cos \c sin \{A -f- B) = cos \G cos \(a — b) 
cos ic cos i(A -f- B) = sin \G cos \(a + b). 



(17) 



146 SPHERICAL TRIGONOMETRY. 

These are commonly known as Gauss's equations, after Gauss, 

who first introduced them into astronomical computations. They 

had, however, been previously published anonymously by Delambre. 

Example. Given a = 132°46'.7, b = 59° SO'.l, C = 56° 28'.4, 

to find the remaining parts. 

a, 132° 46'.7 sin %e sin \{A - B), + 9.719 08 
sin \g cos \{A - B\ + 9.672 34 
tan \(A — B), 0.046 74 
sin \c, 9.847 48 
cos \c sin i(A + B\ 9.850 32 
cos ic cos i(A + B), - 8.715 77 
tan i(A + B), - 1.134 55 
cos \c, 9.851 49 

i(A-B), 48° 4'.6 

§(A + B), 94°11'.7 

Jc, 44°44 / .l 

A, 142° 16'.3 

.£, 46° r.i 

c, 89°28 , .2 

Exercises. 
Compute the remaining three parts of each of the following 
triangles by Gauss's formulae : 

1. A = 79° 28'.6, b = 28° 20'.3, c = 112° 1'.9. 

2. A = 32°58 / .5, j?= 65° 26'.7, o= 56°21'.2. 

3. a = 112° 12'.6, 5 = 124° 48 / .2, (7 = 18° 17 / .0. 

4. a = 52° 22'.2, ^ = 160° 0'.8, C = 129° 52'.4. 

123. Napier's analogies. If, in the preceding problem, only 
the two remaining sides in the one case, or the two remaining 
angles in the other, are wanted, the process may be shortened. 

Dividing the first of (16) by the second, and the third by the 
fourth, we have 

t^i(a + b) = tmic cosi{A + B) ; 

,, \ smi(A — B) 

t™i(a-l>)=tzn i c s . ni{A + B) ; ) 



h 


59 c 


50'.1 s 


a + b, 


192 c 


36'.8 


a — o, 


72 c 


56'.6 


«* + *), 


96° 


18'.4 < 


i(a-b), 


36 c 


28 / .3 c 


iO, 


28° 


14'.2 


sin i(a — 


*)> 


9.77410 


cos • 


W, 


9.944 98 


cos %(a — 


% 


9.905 34 


smi(a + b), + 9.997 37 


sin 


\C,+ 9.674 97 


cos i(a + b\ - 


- 9.040 80 



(18) 



TRANSFORMATION OF THE FORMULAE. 



147 



from which may be found a and b when A, B, and C are given. 
In the same way we find, from (17), 



. , . ™ , ^ sin i(a 

ton HA— B) = cotiC^ 



toni(A+JS) = cot£<7 



sin %(a -f b) ' 
cos | {a — b) 



(19) 



cos i(a -\-b)\ 

from which A and B may be found when a, b, and O are given. 
The equations (18) and (19) are known as Napier's analogies. 



CHAPTER IT. 

MISCELLANEOUS APPLICATIONS. 




124. To find the distance between two points on the earth? s 
surface and their direction from each other, when their latitudes 
and longitudes are given.* 

Let M and N be the two points 
whose latitudes and longitudes are 
given ; P, the pole ; and EQ, the 
equator. Join M to i^by the arc of g( 
a great circle. Also let 
cp = the lat. and A. = the long, of M ; 
<p l = " « V = " " HT. 

If from the pole P we draw 
through M and iV arcs of great circles PMR and PWS, meeting 
the equator in R and S 9 we shall have 

cp = i?J/ = latitude of M ; 
cp' = SN = latitude of JK 

If we suppose P^ to be the meridian from which we reckon 

longitudes, 

Angle QPM = X; 

Angle QPN — V. 

Then because PB = PS = 90°, we have in the triangle MPN 

PM = 90° - <p ; 

PAT = 90° - <p'\ 

Angle P= A - V. 

* It is assumed in the solution of this problem that the earth is a sphere. 
Although the assumption is not strictly correct, the error to which it can give 
rise can never amount to more than a few thousandths of the whole distance. 



MISCELLANEOUS APPLICATIONS. 149 

This is a case, therefore, in which we have given two sides and 
the included angle to find the remaining side. If we put d for 
the required distance, the general formula gives 

cos d = cos PM cos PN -f- sin PM sin PN cos P 



= sin cp sin <p' -f- cos <p cos <p' cos (A — A'). \ 
This equation will suffice to determine the distance between 

the points in arc of a great circle. 

To reduce it to statute miles, the degrees must be multiplied by 

69-J- ; and to reduce to nautical miles, by 60. 

To solve the problem completely we should know not only the distance but 
the direction, or the angle which the great circle joining the two points makes 
with the meridian. This angle is different at the two points, being equal to the 
angle M of the spherical triangle at one point, and to N at the other. Hence 
the complete solution requires the complete solution of the spherical triangle 
PMN, for which we may use the Gaussian equations instead of the form (a). 

Exercises. 

1. Find (1) the distance,* both in degrees and nautical miles, 
between New York and Liverpool, on an arc of a great circle ; 
(2) the direction in which a ship would sail on leaving the one 
port for the other, on an arc of a great circle, and the direction in 
which she would be sailing on approaching her destination. The 
positions of the cities are : 

New York lat. + 40° 42'.7, long. 74° O'.O west. 

Liverpool lat. + 53° 24'.1, long. 2° 59'.1 west. 

2. Compute the distance between Liverpool and Kio de Janeiro, 
the position of the latter being : 

Latitude, - 22° 54'.7. 
Longitude, 43° 9'.0 west. 
Note that the latitude of Rio is algebraically negative, being reckoned south 
from the equator. 

3. If a ship sails from New York, starting due east, and con- 
tinues her course on an arc of a great circle, what will be her 
latitude when she reaches the meridian of Greenwich, and in what 
direction will she then be sailing? 

* By distance, as used here, distance on the arc of a great circle is to be 
understood, unless explicitly stated otherwise. 



150 SPHERICAL TRIGONOMETRY. 

Geometrical Applications. 

125. Def Three straight lines, each perpendicular to the 
other two and all passing through a common point, are called a 
system of rectangular axes. 

Def. The common point of intersection is called the origin. 
Def. Three planes, each perpendicular to the other two, are 
called three rectangular planes. 

Bemarh. It is shown in geometry that three rectangular planes 
intersect each other in lines forming a system of rectangular axes. 

If a sphere have its centre in the common point of intersection 
of three rectangular planes, these planes will intersect its surface 
in three great circles, forming in all eight trirectangular triangles. 

126. Theoeem. The sum of the squares of the cosines of the 
three angles which a straight line forms with three rectangular 
axes is equal to unity. 

This theorem is expressed in trigonometric language thus : 

If ot, /?, and y he the angles which a straight line forms with 
three rectangular axes, then 

cos 2 a -f- cos 2 /3 -|- cos 2 y = 1. (1) 

Proof. Let the line pass through the common point of inter- 
section of the axes. Let a sphere have 
its centre at this point, and let X, Y, 
Z, and P be the points in which the 
axes and the line intersect the spherical 
surface. 

Join PX, PY, and PZ by arcs of 
great circles, and produce ZP until it 
meets XYm Q. Then— 

Because the angles a, /3, and y are formed at the centre of the 

sphere, 

Arc PX measures angle a; 

« PY « " /?; 

" PZ " " y. 

Because XYZ is a trirectangular triangle, Z is the pole of 

X Y, and ZQ is therefore perpendicular to XY. Therefore the 




MISCELLANEOUS APPLICATIONS. 



151 




spherical triangles XQP and YQP are right-angled at Q, and, 

by § 109, 5, 

cos PX = cos a = cos XQ cos QP ; 

cos P Y = cos /? = cos YQ cos $P. 

Taking the squares of these equations and adding them to the 
square of cos PZ = cos y, we have 

cos 2 a: + cos 3 /? + cos V = cos 2 QP (cos 3 XQ + cos 3 YQ) + cos 3 PZ 
= cos 3 QP + cos 2 PZ (because XT = 90°) 
= 1 (because <>Z = 90°). Q.E.D. 
Another proof. Let OX, Y, and 6>Z be the axes meeting 
in 0, and let OP be the line 
making with the lines OX, OY, 
OZ the respective angles 
POX= a; 
POT= /?; 
PO£ = r- 
Through P pass three planes 
parallel to the respective planes 
formed by the axes, taken two and two, namely. 
Plane PSVE \\ plane XOT; 
Plane PQTS \\ plane TOZ; 
Plane PR WQ || plane Z<9X 
These planes will then form a rectangular 
OTQW- VSPP, of which OP will be a diagonal. 
erty of this diagonal (Geom., § 692), 

OP 2 == 6>T 3 + OW 2 + OF 2 . 
Moreover, because these planes are respectively parallel to the 
three rectangular planes, and because each of the axes is perpendic- 
ular to one of these planes, the three planes in question are each 
perpendicular to one of the axes. 

If we join PT, P W, and P V, these lines, being in planes 

which, as just shown, are perpendicular to the axes OX, T, and 

OZ, will be perpendicular to these axes (Geom.), and we shall have 

OT = OP cos POX = OP cos a. l 

OW= OPcosPOY= OPcos/3\ I (b) 

OV = OP cos POZ = OP cos yl J 



parallelopiped 
By the prop- 

(a) 



152 SPHEBICAL TBIGONOMETBY. 

Taking the sum of the squares of these three equations, and 
substituting for OT* + OW % + OV* its value (a), we have 

OP 2 = OP 2 (cos 2 a + cos 2 /? + cos 2 y). 
Dividing by OP 2 , 

cos 2 a -f cos 2 /? + cos 2 y = 1. Q.E.D. (2) 

127. Corollary. It is easily shown that the angles «, /?, and 
^ are the complements of the angles which OP forms with the 
three rectangular planes. For, pass a plane through P and OZ. 
Because OZ 1 plane XO Y, the cutting plane OZP is also perpen- 
dicular to XOY, and the line OQ in which it cuts XOY 1 OZ. 
But, by definition of the inclination of a line to a plane (Geom., 
§ 603), the angle POQ will be the inclination of OP to the plane 
XO Y. Therefore, if we call this inclination a, we have 

a + a = 90° ; 
and in the same way, 

b + /3 = 90° ; 

c + y = 90° ; 

5 and c being the inclinations of the line to the two other planes. 

Therefore sin a = cos a, sin b = cos /?, and sin c = cos y ; whence 

sin 2 # -f- sin 2 5 -\- sin 2 c = 1. (3) 

Because parallel lines have equal inclinations to a plane, the 
angles which any straight line makes with the three planes are 
equal to those made by a parallel to that line through 0. Hence : 

The sum of the squares of the sines of the three angles which 
any straight line makes with three rectangular jplanes is equal to 
unity. 

128. Case of a plane cutting three rectangular planes. "We 
have the following theorem : 

Theorem. If any plane mtersect three rectangular planes, the 
sum of the squares of the cosines of the three angles which it forms 
with them is equal to unity. 

This can be demonstrated from the preceding theorem by 
dropping a perpendicular from the common point of intersection 
of the three rectangular planes upon the cutting plane. Because 
the rectangular axes and this line are each perpendicular to one of 
the four planes, the angles which they form with each other are 



MISCELLANEOUS APPLICATIONS. 



153 




equal to the angles between the planes. The theorem, having 
already been proved for the angles formed 
by the lines, must therefore be true of the 
equal angles formed by the planes. 

But the theorem may also be proved 
independently as follows : 

Pass a sphere around the origin as a 
centre, and let XY, ZZ, ZX be the arcs 
of great circles in which the rectangular 
planes intersect its surface. 

Pass a plane through the centre par- 
allel to the intersecting plane, and let FD be the great circle in 
which it intersects the spherical surface. 

Put 

Angle YDE = a ; \ the three angles which the in- 
Angle XFE = /?; I tersecting plane forms with 
Angle XED = y ; J the three rectangular planes. 

Join XI) by an arc of a great circle. Then — 

Because XT = XZ = 90°, Xis the pole of YZ, and 

XD = 90° ; 

Angle ADC — angle ADB = 90°. 

Because AFD and AFD are triangles of which the side AD 

is 90°, 

cos AFD = cos /? = — cos FAD cos FDA ; 

cos AFD = cos y = — cos FAD cos EDA 

= _ s i n FAD cos EDA 

(because FAD = FAD + EAF = FAD + 90°). 

Taking the sum of the squares, 

cos 2 (3 + cos 3 y = cos 2 ED A 

= 1- sitfEDA. 

Because FDA = ADB - FDD = 90° - a, 

sin 2 EDA = cos 2 a. 
Hence 

cos 2 a + cos 2 fi + cos 2 ^ = 1. Q.E.D. (4) 

129. Corollary. In the same way that we proved the corol- 
lary of the last theorem we may show : 



154 SPHERICAL TRIGONOMETRY. 

If a plane intersects a system of three rectangular axes, form- 
ing with them the respective angles a, b, c, then 

sin 2 a + sin 3 b + sin 2 c = 1. (5) 

Exekcises. 

1. Having a system of three rectangular axes OX, Y, OZ, a 
line OP is drawn from 0, making 

Angle XOP = 60° ; 

Angle TOP = 50°. 
Find the angle ZOP and the angles which OP makes with the 
three planes XOY, YOZ, ZOX. 

2. Supposing OP = 32.965 centimetres, and the angles to have 
the values in the preceding exercise, 

find the lengths of the perpendiculars 
dropped from P upon each of the 
axes OX, OY, and OZ, and the dis- 
tances of the feet of these perpendicu- 
lars from the origin 0. 

3. The same thing being supposed, 
what are the lengths of the respective 

perpendiculars dropped from P upon the three planes XOY, 
YOZ,x&&ZOX% 

4. The same thing being supposed, what are the lengths of the 
projections of OP upon the three planes ? 

130. Methods of defining the direction of a line in space. 
The direction of a line in a plane is defined by the angle which 
it makes with some fixed line in the plane. For example, if we 
have a known fixed line OX, and it is re- 
quired that another line OP through 
shall make an angle -\- 45° with OX, this 
completely fixes the direction of OP. That 
is, there is only one line through which makes an angle of 
-|- 45° with OX, when we employ the method of measuring angles 
defined in Plane Trigonometry, Chap. I. 

But if OP is not confined to one plane, its position is not fixed 




MISCELLANEOUS APPLICATIONS. 



155 




by the angle XOP, because any number of lines may be drawn 
through 6>, some above the plane of the paper and some below it, 
all making the same angle with OX. The student will see that 
these lines are all elements of a cone having as its vertex and 
OX as its axis. 

Hence at least two conditions are necessary to define the direc- 
tion of a line in space. These two quantities may be chosen in 
various ways, of which the following is the most common. 

We have (1) a plane of reference, the position of which we 
suppose fixed, and (2) in this plane we have a fixed line of refer- 
ence OX. We call the plane of 
reference the fundamental plane. 

Let OZ be a line through 0, per- 
pendicular to the plane. 

Let OP be the line of which the 
•direction is to be defined. / " 

From any point P of this line % N 
drop a perpendicular PQ upon the 
fundamental plane, and join OQ. 

The direction of OP is then defined by the following two 
angles : 

(1) The angle POQ which OP forms with its projection OQ; 
that is, the angle between OP and the plane. 

(2) The angle XOQ which the projection of OP makes with OX. 
It will be remarked that the planes of these two angles are 

perpendicular to each other. 

To show that these two angles completely fix the direction of 
OP, we first remark that when the angle XOQ is given the line 
OQ is fixed. 

Next, because PQ is perpendicular to the plane, the point P 
and therefore the line OP must lie in the plane ZOQ, which is 
fixed because its two lines OZ and OQ are fixed. If the angle 
QOP in this (vertical) plane is given, there is only one line OP 
which can form this angle. 

Hence the direction of the line OP is completely determined 
~by the two angles XOQ and QOP. 



156 



SPHERICAL TRIGONOMETRY. 



The plane XOQ is, when used in this way, the fundamental 
plane. 

131. Relation of the preceding system to latitude and longi- 
tude. To form another conception of these two angles, pass a sphere 
around as a centre, and mark on its surface the points and lines 
in which the lines and planes belonging to the preceding figure 
intersect it. Then : 

The fundamental plane OXQ intersects the spherical surface 
in the great circle MXQN. „ 

The line OX intersects it in X. 

The line OQ intersects it in Q. 

The lines OP and OZ intersect / /____, 
it in P and Z. 

¥e therefore have 

Angle XOQ = arc XQ ; 
Angle QOP = arc QP. 

If now we imagine this sphere to 
be the earth, the great circle MN to 
be its equator, Z to be one of its poles, and P any point on its 
surface, then — 

The arc QP or the angle QOP is the latitude of P. 

The arc XQ or angle XOQ is the longitude of P, counted 
from ZX as a prime meridian. 

Thus the angles we have been defining may be described under 
the familiar forms of longitude and latitude. 




133. Position of a pomt. To fix the position of any point 
relative to a fundamental plane, we must select a point O in that 
plane and a line OX as a point and line of reference. If P be 
any point of which we wish to describe the position, we draw 
the line OP and form the same construction as in article 130. 
Then the position of P is fixed by the two angles XOQ and QOP, 
already described, and the distance OP. 

Hence three quantities are required to fix the posiUon of a 
point in space. 



MISCELLANEOUS APPLICATIONS. 



157 



Def The quantities which fix or describe the position of a 
point are called the co-ordinates of the point. 

The angles XOQ and QOP and the length OP are therefore 
co-ordinates of the point P, and are distinguished as polar co- 
ordinates. 

133. Polar distance and longitude. In the preceding figure, 

because MXQN is a great circle, and Z its pole, we have 

Angle XOQ = arc XQ = angle XZP ; 

Arc PQ = 90° - arc ZP. 

Therefore if we put 

p, the arc ZP, which is called the polar distance of P; 

cp, the arc PQ = angle QOP, 

we have 

sin cp = cos^> ; -> 

cos cp = sin p ; L (6) 

A = angle XZP. J 

Thus we may define the direction of a line by polar distance 
and longitude, as well as by latitude and longitude. Applying the 
same system to positions on the earth's surface, their distance from 
the north pole of the earth is used instead of their latitude. Thus 
we should have, 

For New York, p = 49° IT'S; 
For Kio de Janeiro, p = 112° 54Z.7. 



134. Rectangular co-ordinates. The rectangular co-ordinates 
of a point are its distances from 
three rectangular planes. In the 
figure the lines PQ, PR, and PS 
are the rectangular co-ordinates 
of P with respect to the axes 
OZ, OX, and OY respectively. 
In other words : 

The co-ordinate of a point rel- 
ative to each axis is the length of the line parallel to that axis 
from the point to the plane of the two other axes. 




158 



SPHERICAL TRIGONOMETRY. 



We designate these co-ordinates thus : If P is the point whose 
co-ordinates are x, y and z, 

x is the line PP parallel to OX ; 
y " " PS " " OY; 
2 « " PQ « " OZ. 




135. Problem. To find the relation between rectangular 
and polar co-ordinates. 

Let P be the point whose polar 
co-ordinates are given. 

From P drop PQ 1 plane XOY 

From Q drop QE l 6>X. 

"We then have 

QP = z; 

PQ = y; 

OP = x. 

The first of these equations follows immediately from the defi- 
nition of rectangular co-ordinates. 

To see the truth of the second we notice that because 
PQ || plane XOZ, the perpendicular PQ is equal to the perpen- 
dicular from P upon the plane XOZ. 

Let us now put 

cp, the angle QOP, or the elevation of OP, above the plane 
XO Y. We may call this angle the latitude of P. 

X, the angle XOQ which OQ, the projection of OP, makes 
with OX. We may call this angle the longitude of P. 

r, the length of OP. 

We then find 

z = QP = r sin <p ; 

OQ = r cos q>; 

y = PQ = OQ sin A. = r cos <p sin A ; 

a? = 0i? = OQ cos A = y cos <p cos A ; „ 

which are the required expressions. 

If we take the sum of the squares of these three equations, we 
find, by simple reductions, 



(7) 



+Y+*=* 



(8) 



MISCELLANEOUS APPLICATIONS. 159 

136. The second proof of § 120 affords very simple ex- 
pressions for the rectangular co-ordinates, in terms of the angles 
which OP makes with the rectangular axes. 

In the figure we have 

TQ=y;\ (9) 

QP=z. J 
Putting, as before, a, ft, and y for the angles POX, POY, 
and POZ, which OP makes with the axes OX, Y, and OZ 

respectively, we have, by (b), § 126, 

x = r cos a ; -\ 

y = r cos ft ; i- (10) 

z — r cos y. J 
Comparing these with equations (7), we have the following 
values of a, ft, and y in terms of <p and A. : 
cos a = cos cp cos A ; -^ 

cos /3 — cos cp sin X ; J- (11) 

cos y = sin 99. J 

The sum of the squares of the second members is identically 
equal to unity, as it should be. 

137. We may find a linear expression for r in terms of the 
rectangular co-ordinates by multiplying the equations (10) by cos a, 
cos ft, and cos y respectively, and taking the sum of the products, 
noting equation (2). We thus find 

x cos a -\- y cos ft -f- z cos y = r. 

138. The expressions (10) furnish us another definition of the 
rectangular co-ordinates of a point in space, which will sometimes 
be useful. Since OTP is a right angle, the point Tis the projec- 
tion of P upon the axis OX. In the same way W and V are the 
projections of P upon the other axes. Because OT, OW, and 
V are the co-ordinates of P, we have the definition : 

The rectangular co-ordinates of a point are the distances from 
the origin to its projections upon the three axes. 

139. The preceding formulae are directly applicable to posi- 
tions upon the surface of the earth. Let us suppose to be the 



160 



SPHERICAL TRIGONOMETRY. 



centre of the earth ; OZ, its axis, Z being the north pole ; P, a 
point on its surface ; O Y± OX; the great circle XY the equator, 
and the plane XO Y the plane of the equator. 

Also let the great circle ZX be the meridian of Greenwich, 
from which longitudes are counted, and let PQ be the perpen- 
dicular arc of a great circle from P upon the equator. 

The three rectangular planes will 
then be : 

Plane XOY, the plane of the 
earth's equator ; 

Plane ZOX, the plane passing 
through the earth's axis and through 
Greenwich ; 

Plane YOZ, the plane at right 
angles to the other two. 

The angle QOP = cp will then be 
the latitude of the point P, and XOQ = \ will be its longitude. 

The rectangular co-ordinates %, y, and z will be the distances of 
the point P from the respective planes. 

Hence if we put 

p, the radius OP of the earth, 
we shall have 

z = p sin <p, distance from plane of equator ; 

y = p cos cp sin A, distance from plane of Greenwich ; 

x = p cos cp cos A, distance from third plane. 




Exercises. 

1. Assuming the latitude of "Washington to be + 38° 53'.6, its 
longitude west from Greenwich 77° 3'.0, and its distance from the 
earth's centre to be 6369 kilometres, it is required to compute : 

(a) Its distance from the plane of the equator ; 

(h) Its distance from the earth's axis ; 

(c) The circumference of the circle which it describes every 
day in consequence of the earth's rotation on its axis ; 

(d) Its distance from the plane passing through Greenwich and 
the axis of the earth. 



MISCELLANEOUS APPLICATIONS. 161 

2. What is the length of the straight line through the earth 
from Washington to Melbourne ? The position of Melbourne is : 

Latitude, — 37° 3S'.7 ; 
Longitude, 144° 58'. 7 east of Greenwich ; 
Distance, 6369 kilometres from earth's centre. 
Find the angle between the two radii by § 134, and then the distance by § 61. 

3. In the preceding problem, what are the lengths of the seg- 
ments into which the straight line from Washington to Melbourne 
is divided by the plane of the equator ? Begin by finding the dis- 
tance of each point from the plane of the equator. 

4. How far apart are the feet of the respective perpendiculars 
from Washington and Melbourne upon the plane of the equator ? 
And how far is each foot from the centre of the earth ? 

140. Projection of one line upon another. In geometry and 
trigonometry the projection of a finite line a upon an indefinite 
line X is defined as the distance between the feet of the perpen- 
diculars dropped from the ends of a upon X. Hence it has been 
shown that if we put 

p, the projection, 

or, the angle which a makes with X, 
we shall have 

p = a cos a. 

In demonstrating this proposition we have supposed a and X 
to intersect, and therefore to lie in one plane. 

When they are not in one plane a general definition of the 
angle between two such lines is necessary. 

Def. The angle between two non-intersecting lines is the same 
as the angle formed by one of the lines and an intersecting line 
parallel to the other. 

Example. If Xand a do not intersect, the angle between X 
and a is the same as the angle between X and any parallel to a 
intersecting X. 

The theorem which we have proved for a special case is per- 
fectly general, and is as follows : 

If a he the angle between any two lines, a the length of one of 
them, andp the projection of a upon the other, we have 

p = a cos a. 



162 SPHERICAL TRIGONOMETRY. 

Proof. Let BGbe the projected line ; X, the line of projec- 
tion ; BF and GG, the perpendiculars 
from B and G upon X (These per- 
pendiculars are not parallel.) 

Then FG is by definition the pro- 
jection of BG upon X. 

Through F and G pass planes 
each perpendicular to the line X. 

These planes are parallel (Geom., — 
§616). 

The lines FB and G G being perpendicular to X, lie in these 
respective planes (G-eom., § 586). 

Therefore the points B and G lie in these planes. 

Through F draw FH \\ BG, and meeting the plane through 
GG at the point H. 

Because the planes and lines are parallel, 
FH=BG. 

Join HG. Because H and G are in one plane 1 line X, 

JIG l X. 

Therefore 

FG = FH cos HFG; 

or, because FH = a and angle HFG = a, 

FG = a cos a. Q.E.D. 

141. Plane triangles in space. The following problem is of 
constant occurrence in astronomy : 
Given : 

(1) The distance and direction of a point B from some point 
of reference G, the centre of the earth, for example ; 

(2) The distance and direction of another point G from B ; 
Required the distance and direction of G from G. 

It will be remarked that these distances and directions form 
the sides of a plane triangle, of which G, B, and G are the vertices. 

To attack the problem, we may assume G as the origin of co- 
ordinates, and find the rectangular co-ordinates of B and G as 
referred to G. 



MISCELLANEOUS APPLICATIONS. 



163 




Let us suppose 

OX, Y, OZ, the three co-ordinate axes ; 

r, the length OB;* % 

s, the length BO; 

a, /3, y, the angles XOB, 
YOB, and ZOB, which OB makes ^ 
with the three co-ordinate axes ; & 

Bb, Bb', Bb", the perpendicu- 
lars from B upon the three axes ; 

Co, Co', Oc", the perpendiculars from O upon the three axes. 
Then— 

I. Putting x, y, and z for the co-ordinates of B, we have (§ 136) 

x = Ob = r cos a ; -\ 

y = Ob' = r cos ft ; I (12) 

= 0£* = r cos y. J 

II. If we designate the angles between BO and the co-ordinate 
axes by a', ft, and j/, we have (§ 140) 

be = s cos a! ; ^ 
JV = *cos/?'; I (13) 

&V = 5 cos ;/. J 

III. The lines Oc, Oc', Oc" are by definition the co-ordinates 
of the point C. Hence if we put r" =00, and a", /3", y" the 
angles which OC makes with the three co-ordinate axes, we have 



r cos a 
r" cos ft 



= Oc, 
= Oc'; 
= Oc": 



(a) 



t cos y 
or, comparing with (12) and (13), 

r" cos a" = r cos a -\- s cos a! ; 
r" cos /3" = r cos /? + s cos /?' ; 
r" cos x" = r cos 7 + 5 cos ;/. J 
In practice the angles a, j3, y, etc., are not generally conve- 
nient, because they are not independent. We therefore substi- 
tute for them the polar co-ordinates cp and X, denned in § 135, 
putting 



*The lines OB and OC are omitted in the figure in order to avoid complexity. 



164 SPHERICAL TRIGONOMETRY. 

cp and X, the latitude and longitude expressing the direction 
of OB. 

9/, A/, the same for BO. We may conceive cp' to be the angle 
which OB produced makes with the plane of ZZ; or, which is 
the same thing, the angle between BO and a plane through B 
parallel to XY. We also put 

cp", X'\ the corresponding quantities for 00. 
The general equations (a) combined with (11) then give 
r" cos cp" cos X" = r cos cp cos X -J- s cos cp' cos X' ; -\ 
r" cos cp" sin A" = r cos <p sin A. -j- s cos 9/ sin X' ; L (14) 
r" sin cp" = r sin cp -J- 5 sin 9/. J 

When the six quantities which enter into the second members 
are all given, r", cp" ^ and X" may be computed by Prob. VI. Chap. 
IV. But the first two equations may be simplified by the follow- 
ing transformation : 

Multiply the first equation by sin A r , and the second by cos V, 
and subtract the first product from the second. The remainder 
reduces to 

r" cos cp" sin (X" — V) = r cos cp sin (A, — V). 

Now multiply the first by cos A 7 , and the second by sin A', and 
add the products. The sum reduces to 

r" cos cp" cos (X" —X') = r cos cp cos (X — X') -\- s cos cp'. 

From these two equations and the third of (14) the values of 
r", cp'\ and X" — X' may be computed by Prob. YI. Chap. Y. 

We might equally have effected the transformation by using 
sin X and cos X as multipliers, and proceeding exactly as before. 
The equations to be solved would then be 

r" cos cp" sin {X" — X) = s cos q*' sin {V — X) ; ^» 

r" cos cp" cos (X" — X) = r cos <p -\- s cos cp' cos (X' — X) ; I (15) 
r sin cp" = r sin <p -\- s sin cp' m J 






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